answer：Alright, so I'm trying to solve this geometry problem. It says: In triangle ABC, a line BD is drawn such that angle ABD is equal to angle C. We need to find the lengths of segments AD and DC, given that AB is 2 units and AC is 4 units.Hmm, okay. Let me visualize triangle ABC first. Let me sketch it out in my mind. So, triangle ABC with points A, B, and C. AB is 2, AC is 4. BD is a line from B to some point D on AC, I assume, because it's talking about segments AD and DC. So, BD splits AC into AD and DC, which we need to find.Given that angle ABD is equal to angle C. So, angle at B, between AB and BD, is equal to angle at C, which is angle ACB. Interesting.I think I can use the Angle Bisector Theorem here, but wait, is BD an angle bisector? Not exactly, because angle ABD is equal to angle C, not necessarily half of angle ABC. Hmm, so maybe it's a different theorem or maybe similar triangles?Let me think. If angle ABD is equal to angle C, maybe triangles ABD and something else are similar? Let's see.In triangle ABC, angle at C is equal to angle ABD. So, angle ABD = angle C. Also, angle at A is common to both triangle ABC and triangle ABD. Wait, is that true?Wait, triangle ABD has angles at A, B, and D. Triangle ABC has angles at A, B, and C. So, if angle ABD is equal to angle C, and angle at A is common, then maybe triangles ABD and ACB are similar? Let me check.In triangle ABD and triangle ACB:- Angle ABD = angle ACB (given)- Angle at A is common to both trianglesSo, by AA similarity, triangles ABD and ACB are similar.Yes, that makes sense. So, triangle ABD ~ triangle ACB.Therefore, the sides are proportional. So, the ratio of corresponding sides should be equal.So, AB/AC = AD/AB = BD/CB.Given that AB is 2 and AC is 4, so AB/AC = 2/4 = 1/2.Therefore, AD/AB should also be 1/2, which would mean AD = (1/2)*AB = (1/2)*2 = 1.Wait, so AD is 1? Then DC would be AC - AD = 4 - 1 = 3.Is that it? Hmm, seems straightforward, but let me verify.Wait, if triangle ABD ~ triangle ACB, then AB corresponds to AC, BD corresponds to CB, and AD corresponds to AB.So, the ratio is AB/AC = 2/4 = 1/2.Therefore, AD corresponds to AB, so AD = (AB)*(ratio) = 2*(1/2) = 1.Similarly, BD corresponds to CB, so BD = CB*(1/2). But we don't know CB yet.But since we only need AD and DC, and we found AD = 1, so DC = 4 - 1 = 3.Wait, but let me make sure I didn't skip any steps. Maybe I should use the Law of Sines or something else to confirm.In triangle ABC, let's denote angle at A as angle A, angle at B as angle B, and angle at C as angle C.In triangle ABD, angle at A is angle A, angle at B is angle ABD = angle C, so angle at D is 180 - angle A - angle C.Similarly, in triangle ABC, angle at C is angle C, angle at B is angle B, so angle at A is 180 - angle B - angle C.Wait, so in triangle ABD, angle at D is 180 - angle A - angle C, which is equal to angle B in triangle ABC because angle A + angle B + angle C = 180, so angle B = 180 - angle A - angle C.Therefore, angle at D is equal to angle B.So, triangles ABD and ACB have two angles equal, so they are similar.Therefore, the ratio is AB/AC = AD/AB = BD/CB.So, AB = 2, AC = 4, so ratio is 1/2.Therefore, AD = AB*(1/2) = 1, and DC = AC - AD = 3.Alternatively, using the Law of Sines in triangle ABD and triangle ACB.In triangle ABD:AB / sin(angle ADB) = BD / sin(angle A)In triangle ACB:AC / sin(angle ABC) = BC / sin(angle A)But angle ADB is equal to angle ABC, as we saw earlier.So, AB / sin(angle ABC) = BD / sin(angle A)And AC / sin(angle ABC) = BC / sin(angle A)Therefore, AB / AC = BD / BCWhich is consistent with the similarity ratio.So, AB/AC = 1/2, so BD/BC = 1/2, so BD = BC/2.But we don't know BC yet.Alternatively, maybe using the Law of Sines in triangle ABC.In triangle ABC, AB = 2, AC = 4.Law of Sines: AB / sin(angle C) = AC / sin(angle B) = BC / sin(angle A)But angle ABD = angle C, which is angle at B in triangle ABD.Wait, maybe I'm complicating things.Alternatively, since triangles ABD and ACB are similar, the ratio of sides is 1/2, so AD = 1, DC = 3.So, I think that's the answer.**Final Answer**The lengths of segments (AD) and (DC) are (boxed{1}) and (boxed{3}) respectively.

answer：First, I need to determine how many numbers between 1 and 121 are multiples of 3 or 5. To do this, I'll use the principle of inclusion and exclusion.I'll start by finding the number of multiples of 3 within this range. I can calculate this by dividing 121 by 3 and taking the integer part of the result. So, 121 divided by 3 is approximately 40.333, which means there are 40 multiples of 3.Next, I'll find the number of multiples of 5. Similarly, I'll divide 121 by 5, which gives me 24.2. Taking the integer part, there are 24 multiples of 5.However, some numbers are multiples of both 3 and 5, which means they are multiples of 15. To avoid double-counting these numbers, I'll calculate the number of multiples of 15. Dividing 121 by 15 gives approximately 8.066, so there are 8 multiples of 15.Finally, I'll apply the inclusion-exclusion principle by adding the number of multiples of 3 and 5, then subtracting the number of multiples of 15. This gives me 40 + 24 - 8 = 56 numbers that are multiples of 3 or 5 between 1 and 121.

answer：Alright, so I have this problem here: I need to find the largest positive integer ( n ) such that for every prime ( p ) with ( 2 < p < n ), the difference ( n - p ) is also prime. Hmm, okay, let me try to unpack this step by step.First off, let me make sure I understand the problem correctly. I need to find the biggest number ( n ) where, if I take any prime number ( p ) that's bigger than 2 and less than ( n ), then subtracting ( p ) from ( n ) will give me another prime number. So, for example, if ( n ) is 10, and ( p ) is 3, then ( 10 - 3 = 7 ), which is prime. If ( p ) is 5, then ( 10 - 5 = 5 ), which is also prime. But if ( p ) is 7, then ( 10 - 7 = 3 ), which is prime as well. So, 10 seems to satisfy the condition. But is 10 the largest such number? Probably not, because I remember something called the Goldbach conjecture which says that every even number greater than 2 can be expressed as the sum of two primes. If that's the case, maybe larger numbers could satisfy this condition?Wait, hold on. The Goldbach conjecture is about expressing even numbers as the sum of two primes, but this problem is slightly different. It's saying that for every prime ( p ) between 2 and ( n ), ( n - p ) is also prime. So, it's a stronger condition. It's not just that ( n ) can be expressed as the sum of two primes, but that for every prime less than ( n ), subtracting it from ( n ) gives another prime. That seems more restrictive.Let me test this with some numbers. Let's start with ( n = 10 ). The primes between 2 and 10 are 3, 5, 7. Subtracting each from 10:- ( 10 - 3 = 7 ) (prime)- ( 10 - 5 = 5 ) (prime)- ( 10 - 7 = 3 ) (prime)So, 10 works. What about ( n = 12 )? The primes between 2 and 12 are 3, 5, 7, 11.- ( 12 - 3 = 9 ) (not prime)- ( 12 - 5 = 7 ) (prime)- ( 12 - 7 = 5 ) (prime)- ( 12 - 11 = 1 ) (not prime)So, 12 doesn't work because 12 - 3 = 9 is not prime, and 12 - 11 = 1 is not prime either. Therefore, 12 is out.How about ( n = 14 )? Primes between 2 and 14 are 3, 5, 7, 11, 13.- ( 14 - 3 = 11 ) (prime)- ( 14 - 5 = 9 ) (not prime)- ( 14 - 7 = 7 ) (prime)- ( 14 - 11 = 3 ) (prime)- ( 14 - 13 = 1 ) (not prime)Again, 14 fails because 14 - 5 = 9 is not prime and 14 - 13 = 1 isn't prime either.Let me try ( n = 8 ). Primes between 2 and 8 are 3, 5, 7.- ( 8 - 3 = 5 ) (prime)- ( 8 - 5 = 3 ) (prime)- ( 8 - 7 = 1 ) (not prime)So, 8 fails because 8 - 7 = 1 is not prime. Hmm, okay.Wait, so 10 worked, but 8 didn't. What about ( n = 7 )? Primes between 2 and 7 are 3, 5.- ( 7 - 3 = 4 ) (not prime)- ( 7 - 5 = 2 ) (prime)So, 7 fails because 7 - 3 = 4 is not prime.How about ( n = 6 )? Primes between 2 and 6 are 3, 5.- ( 6 - 3 = 3 ) (prime)- ( 6 - 5 = 1 ) (not prime)So, 6 fails as well.Wait, so 10 was the first number that worked. Let's check ( n = 10 ) again.Primes between 2 and 10: 3, 5, 7.- 10 - 3 = 7 (prime)- 10 - 5 = 5 (prime)- 10 - 7 = 3 (prime)All good. So, 10 works. Let's see if 12 is the next even number, but we saw that 12 doesn't work. How about ( n = 14 )? As above, 14 doesn't work. What about 16?Primes between 2 and 16: 3, 5, 7, 11, 13.- 16 - 3 = 13 (prime)- 16 - 5 = 11 (prime)- 16 - 7 = 9 (not prime)- 16 - 11 = 5 (prime)- 16 - 13 = 3 (prime)So, 16 fails because 16 - 7 = 9 is not prime.How about ( n = 18 )?Primes between 2 and 18: 3, 5, 7, 11, 13, 17.- 18 - 3 = 15 (not prime)- 18 - 5 = 13 (prime)- 18 - 7 = 11 (prime)- 18 - 11 = 7 (prime)- 18 - 13 = 5 (prime)- 18 - 17 = 1 (not prime)So, 18 fails because 18 - 3 = 15 and 18 - 17 = 1 are not primes.Hmm, seems like 10 is the only even number so far that works. Let me check ( n = 20 ).Primes between 2 and 20: 3, 5, 7, 11, 13, 17, 19.- 20 - 3 = 17 (prime)- 20 - 5 = 15 (not prime)- 20 - 7 = 13 (prime)- 20 - 11 = 9 (not prime)- 20 - 13 = 7 (prime)- 20 - 17 = 3 (prime)- 20 - 19 = 1 (not prime)So, 20 fails because 20 - 5 = 15, 20 - 11 = 9, and 20 - 19 = 1 are not primes.Wait, so 10 is working, but as I go higher, it's failing. Let me check ( n = 10 ) again and see if maybe there's a larger number that works.Wait, maybe I should think about why 10 works. For ( n = 10 ), the primes less than 10 are 3, 5, 7. Subtracting each from 10 gives 7, 5, 3, which are all primes. So, 10 is special because it's even, and the primes less than it are all odd, so subtracting an odd from an even gives an odd, which can be prime. But when ( n ) increases, the number of primes less than ( n ) increases, so it's more likely that subtracting some prime from ( n ) will result in a composite number.Wait, let me think if there's a way to characterize such numbers ( n ). If ( n ) is such that ( n - p ) is prime for every prime ( p ) with ( 2 < p < n ), then ( n ) must be such that all numbers ( n - p ) are primes. So, ( n ) must be a number where all the numbers ( n - p ) are primes, which is a pretty strict condition.Let me see if I can find any numbers larger than 10 that satisfy this. Let's try ( n = 12 ) again, but as we saw, it fails because 12 - 3 = 9 is not prime. How about ( n = 14 )? 14 - 5 = 9 is not prime. ( n = 16 ): 16 - 7 = 9 is not prime. ( n = 18 ): 18 - 3 = 15 is not prime. ( n = 20 ): 20 - 5 = 15 is not prime. So, seems like 10 is the only even number that works.Wait, but what about odd numbers? Let me check ( n = 7 ). We saw earlier that 7 - 3 = 4 is not prime, so 7 doesn't work. ( n = 9 ): primes between 2 and 9 are 3, 5, 7.- 9 - 3 = 6 (not prime)- 9 - 5 = 4 (not prime)- 9 - 7 = 2 (prime)So, 9 fails because 9 - 3 = 6 and 9 - 5 = 4 are not primes.How about ( n = 11 )? Primes between 2 and 11 are 3, 5, 7.- 11 - 3 = 8 (not prime)- 11 - 5 = 6 (not prime)- 11 - 7 = 4 (not prime)So, 11 fails.Wait, so all the odd numbers I've checked so far don't work because subtracting a prime from an odd number gives an even number, which is only prime if it's 2. So, for example, if ( n ) is odd, then ( n - p ) is even if ( p ) is odd, which all primes greater than 2 are. So, ( n - p ) would have to be 2 to be prime. That means ( p = n - 2 ). But ( p ) is a prime less than ( n ), so ( n - 2 ) must also be prime. But that only gives one prime ( p = n - 2 ), but we need all primes ( p ) with ( 2 < p < n ) to satisfy ( n - p ) is prime. So, for an odd ( n ), unless ( n - p = 2 ) for all primes ( p ), which is impossible unless all primes ( p ) are equal to ( n - 2 ), which is not the case. Therefore, it's impossible for an odd ( n ) to satisfy the condition because subtracting a prime from it would result in an even number, which is only prime if it's 2, but you can't have all primes ( p ) less than ( n ) equal to ( n - 2 ). So, only even numbers can satisfy this condition.Therefore, I can focus on even numbers. So, 10 works, 12 doesn't, 14 doesn't, 16 doesn't, 18 doesn't, 20 doesn't. Let me check ( n = 22 ).Primes between 2 and 22: 3, 5, 7, 11, 13, 17, 19.- 22 - 3 = 19 (prime)- 22 - 5 = 17 (prime)- 22 - 7 = 15 (not prime)- 22 - 11 = 11 (prime)- 22 - 13 = 9 (not prime)- 22 - 17 = 5 (prime)- 22 - 19 = 3 (prime)So, 22 fails because 22 - 7 = 15 and 22 - 13 = 9 are not primes.How about ( n = 24 )?Primes between 2 and 24: 3, 5, 7, 11, 13, 17, 19, 23.- 24 - 3 = 21 (not prime)- 24 - 5 = 19 (prime)- 24 - 7 = 17 (prime)- 24 - 11 = 13 (prime)- 24 - 13 = 11 (prime)- 24 - 17 = 7 (prime)- 24 - 19 = 5 (prime)- 24 - 23 = 1 (not prime)So, 24 fails because 24 - 3 = 21 and 24 - 23 = 1 are not primes.Hmm, seems like 10 is still the only one that works. Let me check ( n = 28 ).Primes between 2 and 28: 3, 5, 7, 11, 13, 17, 19, 23.- 28 - 3 = 25 (not prime)- 28 - 5 = 23 (prime)- 28 - 7 = 21 (not prime)- 28 - 11 = 17 (prime)- 28 - 13 = 15 (not prime)- 28 - 17 = 11 (prime)- 28 - 19 = 9 (not prime)- 28 - 23 = 5 (prime)So, 28 fails because multiple subtractions result in non-primes.Wait, maybe I should think about this differently. If ( n ) is such that all ( n - p ) are primes for primes ( p ) between 2 and ( n ), then ( n ) must be a number where all numbers ( n - p ) are primes. So, ( n ) must be such that ( n - p ) is prime for all primes ( p ) in that range.This seems similar to the concept of a "prime gap" or something related to prime pairs. But I'm not sure. Alternatively, maybe ( n ) is a number where ( n - p ) is prime for all primes ( p ) less than ( n ). So, for ( n = 10 ), we saw that works. Let me see if there's a known concept or theorem about such numbers.Wait, actually, I think these numbers are called "complete numbers" or something similar, but I'm not sure. Maybe I should look for a pattern or try to find a mathematical property that such numbers must satisfy.Let me think about the properties of ( n ). Since ( n ) is even (as we saw earlier, odd numbers can't satisfy the condition because subtracting an odd prime from an odd number gives an even number, which is only prime if it's 2, but that can't happen for all primes ( p )), ( n ) must be even. So, let's focus on even numbers.Now, for ( n ) to satisfy the condition, ( n - p ) must be prime for every prime ( p ) with ( 2 < p < n ). So, for each such prime ( p ), ( n - p ) must be prime. So, ( n ) must be such that it's a prime plus another prime, for every prime less than ( n ).Wait, that sounds a bit like the Goldbach conjecture, which states that every even integer greater than 2 can be expressed as the sum of two primes. But in this case, it's a stronger condition because it's not just that ( n ) can be expressed as the sum of two primes, but that for every prime ( p ) less than ( n ), ( n - p ) is also prime.So, it's not just one pair of primes adding up to ( n ), but all primes ( p ) less than ( n ) must pair with another prime ( q = n - p ) such that ( q ) is also prime.This seems like a very restrictive condition. So, perhaps only small numbers satisfy this.Given that, let me check ( n = 10 ) again. It works. What about ( n = 12 )? It fails because 12 - 3 = 9 is not prime. So, 10 is the largest so far.Wait, but let me check ( n = 14 ) again. 14 - 5 = 9 is not prime, so 14 fails. ( n = 16 ): 16 - 7 = 9 is not prime. ( n = 18 ): 18 - 3 = 15 is not prime. ( n = 20 ): 20 - 5 = 15 is not prime. So, 10 is the only even number that works.But wait, let me think about ( n = 4 ). Primes between 2 and 4 are only 3.- 4 - 3 = 1 (not prime)So, 4 fails.( n = 6 ): primes between 2 and 6 are 3, 5.- 6 - 3 = 3 (prime)- 6 - 5 = 1 (not prime)So, 6 fails.( n = 8 ): primes between 2 and 8 are 3, 5, 7.- 8 - 3 = 5 (prime)- 8 - 5 = 3 (prime)- 8 - 7 = 1 (not prime)So, 8 fails.( n = 10 ): works as we saw.So, 10 is the first even number that works, and the next even numbers fail. So, 10 might be the answer.But wait, let me check ( n = 10 ) again. Are there any primes between 2 and 10 that when subtracted from 10 give a non-prime? Let's list them:Primes between 2 and 10: 3, 5, 7.- 10 - 3 = 7 (prime)- 10 - 5 = 5 (prime)- 10 - 7 = 3 (prime)All are primes. So, 10 works.Is there a larger number that works? Let me think about ( n = 12 ) again. It fails because 12 - 3 = 9 is not prime. Similarly, ( n = 14 ) fails because 14 - 5 = 9 is not prime. So, 10 is the largest.Wait, but let me think about ( n = 24 ). Earlier, I saw that 24 - 3 = 21 is not prime, so it fails. ( n = 30 ):Primes between 2 and 30: 3, 5, 7, 11, 13, 17, 19, 23, 29.- 30 - 3 = 27 (not prime)- 30 - 5 = 25 (not prime)- 30 - 7 = 23 (prime)- 30 - 11 = 19 (prime)- 30 - 13 = 17 (prime)- 30 - 17 = 13 (prime)- 30 - 19 = 11 (prime)- 30 - 23 = 7 (prime)- 30 - 29 = 1 (not prime)So, 30 fails because 30 - 3 = 27, 30 - 5 = 25, and 30 - 29 = 1 are not primes.Therefore, 10 seems to be the only even number that satisfies the condition. So, the largest positive integer ( n ) is 10.Wait, but let me just check ( n = 10 ) one more time to make sure I didn't miss anything. Primes between 2 and 10: 3, 5, 7.- 10 - 3 = 7 (prime)- 10 - 5 = 5 (prime)- 10 - 7 = 3 (prime)Yes, all are primes. So, 10 works. And as we've seen, all larger even numbers fail because subtracting some prime from them results in a composite number.Therefore, the answer is ( boxed{10} ).

answer：Okay, so I have this problem where there are 6 students participating in 4 sports events. The requirements are that students A and B cannot be in the same event. Each event must have at least one participant, and each person can only participate in one event. I need to find the number of different arrangements that meet these requirements. The options are A: 1560, B: 1382, C: 1310, D: 1320.Hmm, let me think about how to approach this. It seems like a combinatorial problem involving permutations and combinations with restrictions.First, without any restrictions, the number of ways to assign 6 students to 4 events where each event has at least one participant is equivalent to the number of onto functions from a set of 6 elements to a set of 4 elements. That can be calculated using the principle of inclusion-exclusion, which gives us the formula:[sum_{k=0}^{4} (-1)^k binom{4}{k} (4 - k)^6]Calculating that:[4^6 - binom{4}{1}3^6 + binom{4}{2}2^6 - binom{4}{3}1^6]Let me compute each term:- (4^6 = 4096)- (binom{4}{1}3^6 = 4 times 729 = 2916)- (binom{4}{2}2^6 = 6 times 64 = 384)- (binom{4}{3}1^6 = 4 times 1 = 4)So, the total is:[4096 - 2916 + 384 - 4 = 4096 - 2916 = 1180; 1180 + 384 = 1564; 1564 - 4 = 1560]So, without any restrictions, the number of arrangements is 1560, which is option A. But we have a restriction: A and B cannot be in the same event. So, we need to subtract the number of arrangements where A and B are in the same event.Alternatively, maybe it's easier to calculate the total number of arrangements where A and B are not in the same event directly.Let me think. If I consider A and B as two separate entities, the total number of arrangements where they are not in the same event can be found by considering the total number of onto functions minus the number of onto functions where A and B are in the same event.So, let's compute the number of arrangements where A and B are in the same event. If A and B are in the same event, we can treat them as a single entity. So, instead of 6 students, we have 5 entities: the AB pair and the other 4 students.Now, we need to assign these 5 entities to 4 events, with each event having at least one participant. So, the number of onto functions from 5 entities to 4 events is:[sum_{k=0}^{4} (-1)^k binom{4}{k} (4 - k)^5]Calculating each term:- (4^5 = 1024)- (binom{4}{1}3^5 = 4 times 243 = 972)- (binom{4}{2}2^5 = 6 times 32 = 192)- (binom{4}{3}1^5 = 4 times 1 = 4)So, the total is:[1024 - 972 + 192 - 4 = 1024 - 972 = 52; 52 + 192 = 244; 244 - 4 = 240]Therefore, the number of arrangements where A and B are in the same event is 240. So, the number of arrangements where A and B are not in the same event is the total number of arrangements without restrictions minus this number.So, 1560 - 240 = 1320.Wait, that gives me 1320, which is option D. But let me verify if this is correct because sometimes when we treat AB as a single entity, we might be overcounting or undercounting.Alternatively, another approach is to consider assigning A first and then assigning B such that B is not in the same event as A.So, let's think about it step by step.First, assign student A to one of the 4 events. There are 4 choices.Then, assign student B to one of the remaining 3 events (since B cannot be in the same event as A). So, 3 choices.Now, assign the remaining 4 students (C, D, E, F) to the 4 events, with the condition that each event must have at least one participant. However, since A is already in one event and B is in another, the remaining 4 students can be assigned to the 4 events without any restrictions except that each event must have at least one participant.Wait, but actually, since A is in one event and B is in another, the other two events might still be empty if all remaining students go to A's or B's events. So, we need to ensure that all 4 events have at least one participant.Therefore, the problem reduces to assigning 4 students to 4 events, with the condition that two specific events (the ones containing A and B) must have at least one more participant each, and the other two events must have at least one participant each.Wait, no. Actually, the events containing A and B already have one participant each, so the remaining 4 students can be assigned to any of the 4 events, but we must ensure that the other two events (not containing A or B) get at least one student each.So, the problem is similar to assigning 4 distinguishable objects (students) into 4 distinguishable boxes (events), with the condition that two specific boxes (the ones not containing A or B) must have at least one object each.So, the number of ways is equal to the total number of assignments minus the assignments where at least one of the two specific boxes is empty.Using inclusion-exclusion again.Total number of assignments without restrictions: (4^4 = 256).Number of assignments where the first specific box (say, event C) is empty: (3^4 = 81).Similarly, number of assignments where the second specific box (event D) is empty: (3^4 = 81).Number of assignments where both specific boxes are empty: (2^4 = 16).Therefore, the number of valid assignments is:[4^4 - binom{2}{1}3^4 + binom{2}{2}2^4 = 256 - 2 times 81 + 16 = 256 - 162 + 16 = 110]So, for each assignment of A and B, we have 110 ways to assign the remaining students.But wait, how many ways are there to assign A and B?We assigned A to 4 events and B to 3 events, so 4 × 3 = 12 ways.Therefore, the total number of arrangements is 12 × 110 = 1320.So, that's the same answer as before, 1320, which is option D.But wait, hold on. Let me think again. Is this correct?Because when we fix A and B in different events, and then assign the remaining students, we have to ensure that all events have at least one participant. Since A is already in one event and B in another, the remaining two events must each have at least one of the remaining students.So, the number of ways to assign the remaining students is equal to the number of onto functions from 4 students to 4 events, with the condition that two specific events (the ones not containing A or B) must have at least one student each.Wait, actually, no. Because the two specific events (C and D) need to have at least one student each, but the other two events (A and B's events) can have more students.So, it's similar to assigning 4 students to 4 events, with the condition that two specific events must have at least one student each.Which is exactly what I calculated earlier, resulting in 110 ways.Therefore, the total number of arrangements is 12 × 110 = 1320.So, that seems consistent.Alternatively, another approach is to consider the total number of onto functions (1560) and subtract the number of onto functions where A and B are in the same event (240), giving 1560 - 240 = 1320.So, both methods lead to the same answer, 1320, which is option D.But wait, let me check if the number of onto functions where A and B are in the same event is indeed 240.Earlier, I treated A and B as a single entity, so 5 entities, and then calculated the number of onto functions as 240. But is that correct?Wait, when we treat A and B as a single entity, we have 5 entities: AB, C, D, E, F.We need to assign these 5 entities to 4 events, with each event having at least one entity. So, the number of onto functions is indeed calculated as:[sum_{k=0}^{4} (-1)^k binom{4}{k} (4 - k)^5 = 4^5 - 4 times 3^5 + 6 times 2^5 - 4 times 1^5]Which is 1024 - 972 + 192 - 4 = 240. So, that's correct.Therefore, subtracting 240 from 1560 gives 1320, which is the number of arrangements where A and B are not in the same event.Therefore, the answer should be D: 1320.But wait, hold on. Let me think again. The problem says "students A and B cannot participate in the same event." So, does that mean that A and B must be in different events? Yes, that's what it means.But in my first approach, I considered all onto functions (1560) and subtracted the ones where A and B are together (240), giving 1320. That seems correct.Alternatively, in the second approach, I considered assigning A and B first, then assigning the rest, leading to 12 × 110 = 1320. So, both methods lead to the same result.Therefore, I think the correct answer is D: 1320.But wait, let me check if the initial total number of onto functions is indeed 1560.Calculating the number of onto functions from 6 students to 4 events is:[sum_{k=0}^{4} (-1)^k binom{4}{k} (4 - k)^6]Which is:4^6 - 4×3^6 + 6×2^6 - 4×1^6Calculating each term:4^6 = 40964×3^6 = 4×729 = 29166×2^6 = 6×64 = 3844×1^6 = 4×1 = 4So, 4096 - 2916 = 11801180 + 384 = 15641564 - 4 = 1560Yes, that's correct.Therefore, 1560 total onto functions.Number of onto functions where A and B are together: 240.Thus, 1560 - 240 = 1320.Therefore, the answer is D: 1320.But wait, let me think if there's another way someone might approach this problem and get a different answer.Suppose someone considers the problem as assigning each student to an event, with A and B not in the same event, and each event must have at least one student.So, the total number of assignments without any restrictions is 4^6 = 4096.But we need to subtract the assignments where A and B are in the same event and also ensure that each event has at least one student.Wait, but that's more complicated because we have two conditions: A and B not in the same event, and each event has at least one student.Alternatively, maybe using inclusion-exclusion for both conditions.But that might be more involved.Alternatively, think of it as two steps: first assign A and B to different events, then assign the remaining students ensuring that all events have at least one student.Which is similar to what I did earlier.So, first assign A: 4 choices.Then assign B: 3 choices.Then assign the remaining 4 students to 4 events, with the condition that the two events not containing A or B must have at least one student each.Which is the same as assigning 4 students to 4 events, with two specific events needing at least one student each.As calculated earlier, that's 110 ways.Therefore, total arrangements: 4 × 3 × 110 = 1320.So, that's consistent.Alternatively, if someone were to think that after assigning A and B, the remaining students can be assigned freely, but then subtract the cases where some events are empty.But that might complicate things.Alternatively, another approach is to calculate the total number of assignments where A and B are in different events and all events have at least one student.Which is exactly what we did.So, I think 1320 is the correct answer, which is option D.Therefore, I think the correct answer is D: 1320.**Final Answer**boxed{D}

answer：Okay, so I have this problem here where I need to find the minimum value of the expression (frac{1}{x^{2}} + frac{4}{y^{2}}) given that (x > 0), (y > 0), and (2x + y = 2). Hmm, let me think about how to approach this.First, since we have a constraint (2x + y = 2), maybe I can express one variable in terms of the other. That way, I can substitute it into the expression I need to minimize and work with a single variable. Let me try that.So, from the equation (2x + y = 2), I can solve for (y):[y = 2 - 2x]Since (y > 0), this means that (2 - 2x > 0), which simplifies to (x < 1). Also, since (x > 0), the domain for (x) is (0 < x < 1).Now, substitute (y = 2 - 2x) into the expression we need to minimize:[frac{1}{x^{2}} + frac{4}{(2 - 2x)^{2}}]Let me simplify that second term. The denominator is ((2 - 2x)^2), which is (4(1 - x)^2). So, the expression becomes:[frac{1}{x^{2}} + frac{4}{4(1 - x)^{2}} = frac{1}{x^{2}} + frac{1}{(1 - x)^{2}}]So now, the problem reduces to minimizing (f(x) = frac{1}{x^{2}} + frac{1}{(1 - x)^{2}}) for (0 < x < 1).Hmm, okay. To find the minimum, I can take the derivative of (f(x)) with respect to (x), set it equal to zero, and solve for (x). Let's compute the derivative.First, rewrite (f(x)) as:[f(x) = x^{-2} + (1 - x)^{-2}]Now, take the derivative term by term:- The derivative of (x^{-2}) is (-2x^{-3}).- The derivative of ((1 - x)^{-2}) is (2(1 - x)^{-3}) because of the chain rule (the derivative of (1 - x) is (-1), so the negative cancels out).So, putting it together:[f'(x) = -2x^{-3} + 2(1 - x)^{-3}]Simplify:[f'(x) = -frac{2}{x^{3}} + frac{2}{(1 - x)^{3}}]To find critical points, set (f'(x) = 0):[-frac{2}{x^{3}} + frac{2}{(1 - x)^{3}} = 0]Let me move one term to the other side:[frac{2}{(1 - x)^{3}} = frac{2}{x^{3}}]Divide both sides by 2:[frac{1}{(1 - x)^{3}} = frac{1}{x^{3}}]Take reciprocals on both sides:[(1 - x)^{3} = x^{3}]Hmm, so ((1 - x)^3 = x^3). Let me take the cube root of both sides:[1 - x = x]Wait, is that right? Because cube root of (x^3) is (x), and cube root of ((1 - x)^3) is (1 - x). So, yes, that gives:[1 - x = x]Solving for (x):[1 = 2x implies x = frac{1}{2}]Okay, so (x = frac{1}{2}) is a critical point. Now, I need to check if this is a minimum.First, let's check the second derivative to confirm concavity. Alternatively, I can test points around (x = frac{1}{2}) to see if the function is decreasing before and increasing after, which would indicate a minimum.But maybe computing the second derivative is more straightforward. Let me try that.First, recall that the first derivative is:[f'(x) = -frac{2}{x^{3}} + frac{2}{(1 - x)^{3}}]So, the second derivative (f''(x)) is the derivative of (f'(x)):- The derivative of (-2x^{-3}) is (6x^{-4}).- The derivative of (2(1 - x)^{-3}) is (6(1 - x)^{-4}) because of the chain rule (derivative of (1 - x) is (-1), so the negative cancels out).So, putting it together:[f''(x) = 6x^{-4} + 6(1 - x)^{-4} = frac{6}{x^{4}} + frac{6}{(1 - x)^{4}}]Since (x) is between 0 and 1, both terms are positive, so (f''(x) > 0). This means the function is concave upward at (x = frac{1}{2}), so this critical point is indeed a local minimum. Given the behavior of the function as (x) approaches 0 or 1, the function tends to infinity, so this local minimum must be the global minimum.Therefore, the minimum occurs at (x = frac{1}{2}). Let's find the corresponding (y) using the constraint (2x + y = 2):[2left(frac{1}{2}right) + y = 2 implies 1 + y = 2 implies y = 1]So, (x = frac{1}{2}) and (y = 1). Now, let's compute the value of the expression at this point:[frac{1}{x^{2}} + frac{4}{y^{2}} = frac{1}{left(frac{1}{2}right)^2} + frac{4}{1^2} = frac{1}{frac{1}{4}} + 4 = 4 + 4 = 8]So, the minimum value is 8.Wait, let me double-check my calculations. So, substituting (x = frac{1}{2}) into (frac{1}{x^2}) gives (frac{1}{(frac{1}{2})^2} = frac{1}{frac{1}{4}} = 4). Then, substituting (y = 1) into (frac{4}{y^2}) gives (frac{4}{1} = 4). So, 4 + 4 is indeed 8. That seems correct.Alternatively, maybe I can approach this problem using the method of Lagrange multipliers, just to confirm. Let me try that.Let me define the function to minimize as:[f(x, y) = frac{1}{x^{2}} + frac{4}{y^{2}}]And the constraint is:[g(x, y) = 2x + y - 2 = 0]The method of Lagrange multipliers tells us that at the minimum, the gradient of (f) is proportional to the gradient of (g). So, there exists a (lambda) such that:[nabla f = lambda nabla g]Compute the gradients.First, the gradient of (f):[nabla f = left( frac{partial f}{partial x}, frac{partial f}{partial y} right)]Compute partial derivatives:- (frac{partial f}{partial x} = -2x^{-3})- (frac{partial f}{partial y} = -8y^{-3})Gradient of (g):[nabla g = (2, 1)]So, setting up the equations:[-2x^{-3} = lambda cdot 2 quad (1)][-8y^{-3} = lambda cdot 1 quad (2)]From equation (1):[lambda = frac{-2x^{-3}}{2} = -x^{-3}]From equation (2):[lambda = -8y^{-3}]Set them equal:[-x^{-3} = -8y^{-3}]Multiply both sides by -1:[x^{-3} = 8y^{-3}]Which can be written as:[frac{1}{x^{3}} = frac{8}{y^{3}}]Cross-multiplying:[y^{3} = 8x^{3}]Taking cube roots:[y = 2x]So, we have (y = 2x). Now, substitute this into the constraint (2x + y = 2):[2x + 2x = 2 implies 4x = 2 implies x = frac{1}{2}]Then, (y = 2x = 2 times frac{1}{2} = 1). So, same result as before. Therefore, the minimum occurs at (x = frac{1}{2}), (y = 1), and the minimum value is 8.Hmm, so both methods give the same result. That gives me more confidence that the answer is correct.Alternatively, maybe I can use the Cauchy-Schwarz inequality or AM-GM inequality to solve this. Let me see if that's possible.Looking at the expression (frac{1}{x^{2}} + frac{4}{y^{2}}), it's a sum of reciprocals squared. Maybe I can apply Cauchy-Schwarz.Recall that Cauchy-Schwarz inequality states that for vectors (a) and (b), ((a cdot b)^2 leq (a cdot a)(b cdot b)).Alternatively, in another form, for positive real numbers, ((u_1^2 + u_2^2)(v_1^2 + v_2^2) geq (u_1v_1 + u_2v_2)^2).But I'm not sure if that directly applies here. Alternatively, maybe I can use the method of substitution.Wait, another idea: Maybe use substitution with the constraint.We have (2x + y = 2), so (y = 2 - 2x). Then, as I did earlier, substitute into the expression.But I already did that, so perhaps another approach is to use substitution with variables.Wait, another thought: Maybe set (a = x), (b = y/2), so that the constraint becomes (2a + 2b = 2), which simplifies to (a + b = 1). Then, the expression becomes (frac{1}{a^2} + frac{4}{(2b)^2} = frac{1}{a^2} + frac{4}{4b^2} = frac{1}{a^2} + frac{1}{b^2}).So, now, the problem is to minimize (frac{1}{a^2} + frac{1}{b^2}) with (a + b = 1), (a > 0), (b > 0).Hmm, that seems symmetric. Maybe I can apply some inequality here.Wait, if (a + b = 1), then perhaps using the Cauchy-Schwarz inequality on the terms (frac{1}{a^2}) and (frac{1}{b^2}).Let me consider vectors ((1, 1)) and ((frac{1}{a}, frac{1}{b})). Then, by Cauchy-Schwarz:[left(1 cdot frac{1}{a} + 1 cdot frac{1}{b}right)^2 leq left(1^2 + 1^2right)left(left(frac{1}{a}right)^2 + left(frac{1}{b}right)^2right)]Simplify:[left(frac{1}{a} + frac{1}{b}right)^2 leq 2left(frac{1}{a^2} + frac{1}{b^2}right)]So, (frac{1}{a^2} + frac{1}{b^2} geq frac{1}{2}left(frac{1}{a} + frac{1}{b}right)^2).But I'm not sure if that helps me find the minimum. Alternatively, maybe I can use the method of Lagrange multipliers again, but in terms of (a) and (b).Wait, perhaps another substitution: Let me set (a = t), so (b = 1 - t), with (0 < t < 1). Then, the expression becomes:[frac{1}{t^2} + frac{1}{(1 - t)^2}]Which is exactly the same as the function (f(x)) I had earlier. So, it's the same problem.Alternatively, maybe use the AM-GM inequality. Let me recall that for positive numbers, the arithmetic mean is greater than or equal to the geometric mean.But I have two terms: (frac{1}{x^2}) and (frac{4}{y^2}). Maybe I can write them as squares and apply AM-GM.Wait, another idea: Let me consider the variables (u = frac{1}{x}) and (v = frac{2}{y}). Then, the expression becomes (u^2 + v^2). The constraint (2x + y = 2) can be rewritten in terms of (u) and (v).From (u = frac{1}{x}), we get (x = frac{1}{u}). From (v = frac{2}{y}), we get (y = frac{2}{v}). Substitute into the constraint:[2left(frac{1}{u}right) + frac{2}{v} = 2]Simplify:[frac{2}{u} + frac{2}{v} = 2]Divide both sides by 2:[frac{1}{u} + frac{1}{v} = 1]So, now, we have to minimize (u^2 + v^2) subject to (frac{1}{u} + frac{1}{v} = 1), where (u > 0), (v > 0).Hmm, this seems like a different problem, but maybe it's easier to handle. Let me try to use Cauchy-Schwarz here.We can write:[(u^2 + v^2)left(left(frac{1}{u}right)^2 + left(frac{1}{v}right)^2right) geq left(u cdot frac{1}{u} + v cdot frac{1}{v}right)^2 = (1 + 1)^2 = 4]So,[(u^2 + v^2)left(frac{1}{u^2} + frac{1}{v^2}right) geq 4]But from the constraint, (frac{1}{u} + frac{1}{v} = 1). Let me denote (S = frac{1}{u} + frac{1}{v} = 1), and (P = frac{1}{u} cdot frac{1}{v}). Then, (frac{1}{u^2} + frac{1}{v^2} = S^2 - 2P = 1 - 2P).So, substituting back into the inequality:[(u^2 + v^2)(1 - 2P) geq 4]But I don't know what (P) is. Alternatively, maybe another approach.Let me express (v) in terms of (u) from the constraint:[frac{1}{u} + frac{1}{v} = 1 implies frac{1}{v} = 1 - frac{1}{u} implies v = frac{1}{1 - frac{1}{u}} = frac{u}{u - 1}]So, (v = frac{u}{u - 1}). Now, substitute this into (u^2 + v^2):[u^2 + left(frac{u}{u - 1}right)^2]Let me compute this expression:[u^2 + frac{u^2}{(u - 1)^2}]Let me denote (t = u), so the expression becomes:[t^2 + frac{t^2}{(t - 1)^2}]To find the minimum, take the derivative with respect to (t):First, rewrite the expression:[f(t) = t^2 + frac{t^2}{(t - 1)^2}]Compute the derivative:[f'(t) = 2t + frac{2t(t - 1)^2 - t^2 cdot 2(t - 1)}{(t - 1)^4}]Simplify the numerator of the second term:[2t(t - 1)^2 - 2t^2(t - 1) = 2t(t - 1)[(t - 1) - t] = 2t(t - 1)(-1) = -2t(t - 1)]So, the derivative becomes:[f'(t) = 2t + frac{-2t(t - 1)}{(t - 1)^4} = 2t - frac{2t}{(t - 1)^3}]Set (f'(t) = 0):[2t - frac{2t}{(t - 1)^3} = 0]Factor out (2t):[2tleft(1 - frac{1}{(t - 1)^3}right) = 0]Since (t > 1) (because (v = frac{u}{u - 1}) must be positive, so (u > 1)), (2t neq 0). Therefore:[1 - frac{1}{(t - 1)^3} = 0 implies frac{1}{(t - 1)^3} = 1 implies (t - 1)^3 = 1 implies t - 1 = 1 implies t = 2]So, (t = 2), which means (u = 2). Then, from the constraint, (v = frac{u}{u - 1} = frac{2}{1} = 2).Therefore, (u = 2), (v = 2). Then, the expression (u^2 + v^2 = 4 + 4 = 8). So, same result as before.So, regardless of the method I use—whether substitution, Lagrange multipliers, or even variable substitution with Cauchy-Schwarz—I end up with the minimum value being 8.Just to make sure, let me test a point near (x = frac{1}{2}) to see if the function is indeed higher there. Let's take (x = 0.4), so (y = 2 - 2(0.4) = 1.2). Then, compute the expression:[frac{1}{0.4^2} + frac{4}{1.2^2} = frac{1}{0.16} + frac{4}{1.44} = 6.25 + 2.777... approx 9.027]Which is higher than 8.Similarly, take (x = 0.6), so (y = 2 - 2(0.6) = 0.8). Compute the expression:[frac{1}{0.6^2} + frac{4}{0.8^2} = frac{1}{0.36} + frac{4}{0.64} approx 2.777... + 6.25 approx 9.027]Again, higher than 8.So, this numerical check also supports that 8 is indeed the minimum value.Therefore, after multiple approaches and checks, I'm confident that the minimum value of (frac{1}{x^{2}} + frac{4}{y^{2}}) under the given constraints is 8.**Final Answer**The minimum value is boxed{8}.

answer：First, I need to understand the given ratio of students in the first, second, and third grades, which is 3:3:4. This means that for every 3 students in the first grade, there are 3 in the second grade and 4 in the third grade.Next, I'll calculate the total number of parts in the ratio by adding them together: 3 + 3 + 4, which equals 10 parts.Since the total sample size is 50 students, each part of the ratio corresponds to 50 divided by 10, which is 5 students per part.Finally, to find out how many students should be drawn from the third grade, I'll multiply the number of parts for the third grade (which is 4) by the number of students per part: 4 times 5 equals 20 students.