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# Mrs. Smith has been given film vouchers. Each voucher allows

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Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]  29 Jul 2010, 13:24
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Jan 2013, 02:16, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Voucher!!! [#permalink]  29 Jul 2010, 14:06
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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=n-8 or a+b+c+d=m with m=n-8
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=m-1), ..., (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*(m-1)+...+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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Re: Voucher!!! [#permalink]  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=n-8 or a+b+c+d=m with m=n-8
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=m-1), ..., (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*(m-1)+...+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]  29 Jul 2010, 17:41
9
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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=n-8 or a+b+c+d=m with m=n-8
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=m-1), ..., (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*(m-1)+...+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+r-1Cr-1.
Here if each nephew gets at least 2 tickets we have n-8 tickets left to distribute among 4 nephews and each one of them can receive 0,1,2..n-8 items.
Total no of ways that can happen = (n-8)+(4-1)C(4-1)
= n-5C3
n-5C3=120 for n=15 .So C
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Re: Voucher!!! [#permalink]  29 Jul 2010, 18:42
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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 $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ 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,

$$|t|ttt|t$$
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 $$(5+4-1)C(4-1)=8C3$$.

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 $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

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 $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.
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Re: Voucher!!! [#permalink]  29 Jul 2010, 18:57
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 $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ 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,

$$|t|ttt|t$$
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 $$(5+4-1)C(4-1)=8C3$$.

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 $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

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 $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

Hope it helps.

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Re: Voucher!!! [#permalink]  02 Jul 2011, 17:53
5 stars for the "stars and bars" method! The coolest trick ever!
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Re: Voucher!!! [#permalink]  06 Jul 2011, 10:22
really great trick!!
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Re: Voucher!!! [#permalink]  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]  07 Jan 2013, 11:04
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+r-1Cr-1.

Thanks a lot
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Re: Voucher!!! [#permalink]  07 Jan 2013, 20:10
3
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Expert's post
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 ... 93-part-1/

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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Re: Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]  22 Jan 2014, 15:12
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Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]  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 $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ 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,

$$|t|ttt|t$$
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 $$(5+4-1)C(4-1)=8C3$$.

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 $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

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 $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

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 $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$
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Re: Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]  20 Oct 2014, 13:10
1
KUDOS
Expert's post
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 $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ 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,

$$|t|ttt|t$$
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 $$(5+4-1)C(4-1)=8C3$$.

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 $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$. --> $$k=7$$. Plus the 8 tickets we booked earlier: $$x=k+8=7+8=15$$.

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 $$n+r-1C_{r-1}$$.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is $$n-1C_{r-1}$$.

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 $$(K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120$$.

$$(k+1)(k+2)(k+3)=3!*120=720$$

(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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Re: Mrs. Smith has been given film vouchers. Each voucher allows [#permalink]  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.
Re: Mrs. Smith has been given film vouchers. Each voucher allows   [#permalink] 20 Oct 2014, 14:00
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