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Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]
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Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
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It took me 10 minutes to solve it First, it is remarked that all vouchers are equivalent. Let's call nephews A, B, C, D and their number of vouchers (2+a), (2+b), (2+c), and (2+d), respectively, where a, b, c, d non negative integers (since each of them has at least 2 vouchers). If n is the total number of vouchers, then: a+b+c+d=n8 or a+b+c+d=m with m=n8 Let X=a+b and Y=c+d so X+Y=m and (m+1) possible couples are (X=0, Y=m), (X=1, Y=m1), ..., (X=m, Y=0) On the other hand, (a, b) such that a+b=k and a,b>=0, the number of such couples is equal to k+1. Therefore, generic solution is: 1*(m+1)+2*m+3*(m1)+...+m*2+(m+1)*1 To get 120 different ways, m equals to 7, or n=15, thus C. Sure this is not the easiest way to deal with this problem!
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29 Jul 2010, 15:04
bogos wrote: It took me 10 minutes to solve it First, it is remarked that all vouchers are equivalent. Let's call nephews A, B, C, D and their number of vouchers (2+a), (2+b), (2+c), and (2+d), respectively, where a, b, c, d non negative integers (since each of them has at least 2 vouchers). If n is the total number of vouchers, then: a+b+c+d=n8 or a+b+c+d=m with m=n8 Let X=a+b and Y=c+d so X+Y=m and (m+1) possible couples are (X=0, Y=m), (X=1, Y=m1), ..., (X=m, Y=0) On the other hand, (a, b) such that a+b=k and a,b>=0, the number of such couples is equal to k+1. Therefore, generic solution is: 1*(m+1)+2*m+3*(m1)+...+m*2+(m+1)*1 To get 120 different ways, m equals to 7, or n=15, thus C. Sure this is not the easiest way to deal with this problem! Anyway, this is a good way to solve this tough problem! Thanks



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Re: Voucher!!! [#permalink]
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bogos wrote: It took me 10 minutes to solve it First, it is remarked that all vouchers are equivalent. Let's call nephews A, B, C, D and their number of vouchers (2+a), (2+b), (2+c), and (2+d), respectively, where a, b, c, d non negative integers (since each of them has at least 2 vouchers). If n is the total number of vouchers, then: a+b+c+d=n8 or a+b+c+d=m with m=n8 Let X=a+b and Y=c+d so X+Y=m and (m+1) possible couples are (X=0, Y=m), (X=1, Y=m1), ..., (X=m, Y=0) On the other hand, (a, b) such that a+b=k and a,b>=0, the number of such couples is equal to k+1. Therefore, generic solution is: 1*(m+1)+2*m+3*(m1)+...+m*2+(m+1)*1 To get 120 different ways, m equals to 7, or n=15, thus C. Sure this is not the easiest way to deal with this problem! Actually there is a formula for this The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items = n+r1Cr1. Here if each nephew gets at least 2 tickets we have n8 tickets left to distribute among 4 nephews and each one of them can receive 0,1,2..n8 items. Total no of ways that can happen = (n8)+(41)C(41) = n5C3 n5C3=120 for n=15 .So C
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dungtd wrote: Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
Could someone help me out? Thanks a lot!!! Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x8\) vouchers, so that each can get from zero to \(x8\) as at "least 2", or 2*4=8, we already booked. Let \(x8\) be \(k\). In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept. Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions). Consider: \(ttttt\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets: \(ttttt\) Means that first nephew will get all the tickets, \(ttttt\) Means that first got 0, second 1, third 3, and fourth 1 And so on. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\). So, # of ways to distribute 5 tickets among 4 people is \((5+41)C(41)=8C3\). For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\). \((k+1)(k+2)(k+3)=3!*120=720\). > \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\). Answer: C (15). P.S. How to solve \((k+1)(k+2)(k+3)=3!*120=720\): 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try \(5*6*7=210<720\), next possible triplet \(8*9*10=720\), OK. So \(k+1=8\) > \(k=7\). P.P.S. Direct formula: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).Hope it helps.
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Bunuel wrote: dungtd wrote: Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
Could someone help me out? Thanks a lot!!! Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x8\) vouchers, so that each can get from zero to \(x8\) as at "least 2", or 2*4=8, we already booked. Let \(x8\) be \(k\). In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept. Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions). Consider: \(ttttt\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets: \(ttttt\) Means that first nephew will get all the tickets, \(ttttt\) Means that first got 0, second 1, third 3, and fourth 1 And so on. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\). So, # of ways to distribute 5 tickets among 4 people is \((5+41)C(41)=8C3\). For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\). \((k+1)(k+2)(k+3)=3!*120=720\). > \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\). Answer: C (15). P.S. How to solve \((k+1)(k+2)(k+3)=3!*120=720\): 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try \(5*6*7=210<720\), next possible triplet \(8*9*10=720\), OK. So \(k+1=8\) > \(k=7\). P.P.S. Direct formula: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).Hope it helps. Thanks for your great explanation!



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Re: Voucher!!! [#permalink]
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02 Jul 2011, 17:53
5 stars for the "stars and bars" method! The coolest trick ever!



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really great trick!!



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Re: Voucher!!! [#permalink]
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23 Jul 2011, 22:52
Hi Bunuel,
Why do we assume k=5?
What would happen if we assume some other value of k?
Thanks!



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Re: Voucher!!! [#permalink]
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this formula is worth remembering: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items = n+r1Cr1.
Thanks a lot



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Re: Voucher!!! [#permalink]
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dungtd wrote: Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
Could someone help me out? Thanks a lot!!! This question is a spin on 'distribute n identical things among m people (some people may get none)' Since she must give each of the 4 nephews atleast 2 vouchers, she must have atleast 8. But she can split the vouchers in 120 ways so she certainly has more than 8. Let's assume she gives 2 to each nephew and has the leftover vouchers in her hand. This becomes your standard 'distribute n identical things among m people (some people may get none)' question. Assume the answer is 15 (it's in the middle) and she has 15  8 = 7 vouchers left in her hand. In how many ways can she distribute them among 4 people? In 10!/7!*3! = 10*9*8/(3*2) = 120 ways. This is the required answer so (C) This concept is explained in detail here: http://www.veritasprep.com/blog/2011/12 ... 93part1/Check method II or question 2 on the link above. Note: Say, you had obtained a number less than 120. You would have assumed 16 next and repeated the calculation. Had you obtained a number more than 120, you would have assumed 13 or 14 and done the calculation. 2 iterations would have certainly given you the answer.
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Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]
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20 Oct 2014, 12:01
Bunuel wrote: dungtd wrote: Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
Could someone help me out? Thanks a lot!!! Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x8\) vouchers, so that each can get from zero to \(x8\) as at "least 2", or 2*4=8, we already booked. Let \(x8\) be \(k\). In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept. Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions). Consider: \(ttttt\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets: \(ttttt\) Means that first nephew will get all the tickets, \(ttttt\) Means that first got 0, second 1, third 3, and fourth 1 And so on. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\). So, # of ways to distribute 5 tickets among 4 people is \((5+41)C(41)=8C3\). For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\). \((k+1)(k+2)(k+3)=3!*120=720\). > \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\). Answer: C (15). P.S. How to solve \((k+1)(k+2)(k+3)=3!*120=720\): 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try \(5*6*7=210<720\), next possible triplet \(8*9*10=720\), OK. So \(k+1=8\) > \(k=7\). P.P.S. Direct formula: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).Hope it helps. Bunuel, Blast from the past, but I'm just a bit unclear about the rearrangement here, and why/how the k! cancels out when we multiply by the 3! For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\). \((k+1)(k+2)(k+3)=3!*120=720\)



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Re: Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]
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20 Oct 2014, 13:10
bsmith37 wrote: Bunuel wrote: dungtd wrote: Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
Could someone help me out? Thanks a lot!!! Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x8\) vouchers, so that each can get from zero to \(x8\) as at "least 2", or 2*4=8, we already booked. Let \(x8\) be \(k\). In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept. Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions). Consider: \(ttttt\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets: \(ttttt\) Means that first nephew will get all the tickets, \(ttttt\) Means that first got 0, second 1, third 3, and fourth 1 And so on. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\). So, # of ways to distribute 5 tickets among 4 people is \((5+41)C(41)=8C3\). For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\). \((k+1)(k+2)(k+3)=3!*120=720\). > \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\). Answer: C (15). P.S. How to solve \((k+1)(k+2)(k+3)=3!*120=720\): 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try \(5*6*7=210<720\), next possible triplet \(8*9*10=720\), OK. So \(k+1=8\) > \(k=7\). P.P.S. Direct formula: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).Hope it helps. Bunuel, Blast from the past, but I'm just a bit unclear about the rearrangement here, and why/how the k! cancels out when we multiply by the 3! For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+41)C(41)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\). \((k+1)(k+2)(k+3)=3!*120=720\) (k+3)! = k!*(k+1)(k+2)(k+3), so \(\frac{(k+3)!}{k!3!}=\frac{k!*(k+1)(k+2)(k+3)}{k!3!}=\frac{(k+1)(k+2)(k+3)}{3!}\).
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20 Oct 2014, 14:00
Quote: (k+3)! = k!*(k+1)(k+2)(k+3), so \(\frac{(k+3)!}{k!3!}=\frac{k!*(k+1)(k+2)(k+3)}{k!3!}=\frac{(k+1)(k+2)(k+3)}{3!}\).
that is what i figured. thank you.



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26 Oct 2015, 21:51
Let us assume the total number of vouchers is 15. Each nephew should get at least 2 vouchers. The distribution can contain three 2 vouchers, two 2 vouchers , one two voucher or no 2 voucher. For three 2 vouchers the possibilities are 2 2 2 9, 2 2 9 2, 2 9 2 2 and 9 2 2 2 . So there are 4 possibilities. For two 2 vouchers, one 2 voucher and no two voucher, the possibilities can be shown to be 36, 60 and 20. So the total number of possibilities are 120 as required. Even if we did not choose 15 as the number of vouchers we can arrive at it with a couple of choices.
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26 Dec 2015, 08:45
I was wondering does the question mean 1. each nephew gets 2 vouchers first, 2. the remaining vouchers can be distributed in 120 ways.
so each gets 2 first. 2*4=8 then the remaining vouchers can be distributed in 120 ways (K+3)! / K!*3! = 120 (K+3) (K+2) (K+1) = 720 plug in numbers K = 7 7+ the original 8 = 15. is there anything wrong with my understanding ? thanks



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Re: Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]
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29 Dec 2015, 01:41
VeritasPrepKarishma wrote: dungtd wrote: Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16
Could someone help me out? Thanks a lot!!! This question is a spin on 'distribute n identical things among m people (some people may get none)' Since she must give each of the 4 nephews atleast 2 vouchers, she must have atleast 8. But she can split the vouchers in 120 ways so she certainly has more than 8. Let's assume she gives 2 to each nephew and has the leftover vouchers in her hand. This becomes your standard 'distribute n identical things among m people (some people may get none)' question. Assume the answer is 15 (it's in the middle) and she has 15  8 = 7 vouchers left in her hand. In how many ways can she distribute them among 4 people? In 10!/7!*3! = 10*9*8/(3*2) = 120 ways. This is the required answer so (C) This concept is explained in detail here: http://www.veritasprep.com/blog/2011/12 ... 93part1/Check method II or question 2 on the link above. Note: Say, you had obtained a number less than 120. You would have assumed 16 next and repeated the calculation. Had you obtained a number more than 120, you would have assumed 13 or 14 and done the calculation. 2 iterations would have certainly given you the answer. Quote: I have one confusion though. So, when we distribute the remaining 7 vouchers we are dividing by 7! and 3!. These take care of the arrangements with 7 identical vouchers and 3 identical bars (vvvvvvv), right?. But what about arrangements like vvvvvvv? Here, how do we take care of the identical arrangements with vvv, vv? If we do not take care of them, then with vvv, there will be additional 3! = 6 arrangements multiplied by 2! arranments for vv, right?
Can you please clarify? I have this confusion for other similar porblems. Maybe I am missing some point here.
Consider this question: How many words can you make using all letters of GOSSIP? This is 6!/2! How about number of words using all letters of MMIISS? This is 6!/2!*2!*2! (this includes all words such as MIISMS, ISMISM, ISSMMI etc) On the same lines, consider VVVVVVVIII  Is it any different from above? This would be 10!/7!*3! (this would include all words such as VVVIVIVVIV) We have already taken care of all identical Vs when we divide by 7!
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