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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Mrs. Smith has been given film vouchers. Each voucher allows

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Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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Debashis Roy wrote:
Why dont we consider the no of ways in which 8 vouchers are to be distributed among the 4 nephews equally? Isnt that also included within 120 ways?

Note that the vouchers are not distinct. Since each nephew gets at least 2 vouchers, we give away 2 to each in 1 way. The 8 vouchers are distributed among 4 nephews in 1 way only (2 vouchers per nephew).
The extra 7 vouchers (since we have a total of 15) need to be distributed among the 4 nephews such that any nephew may get any number of vouchers. This can be done in
10!/7!*3! = 10*3*4 = 120 ways

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Karishma
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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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Crack700 pls how did n-5C3=120 give 15? How did you solve for n from that formular to give 15??? Can anyone help? I'm lost please.

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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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[quote="crack700"][quote="bogos"]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![/quote]

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[/quote]
Please I'm confused as to how you got 15 when you solved n-5C3=120

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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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Same question as last post

Using the direct nCr formula does not equate to 120 ...
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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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Why are we assuming that 120 ways to distribute the remaining vouchers, after Mrs Smith has already distributed 2 vouchers per head?
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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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PriyankaPalit7 wrote:

Why are we assuming that 120 ways to distribute the remaining vouchers, after Mrs Smith has already distributed 2 vouchers per head?

Hi
The info in question is
There are 4 nephews.
Each has to be given 2 or more vouchers.
If there are x vouchers, these x can be distributed in 120 ways.
We have to find x.
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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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PriyankaPalit7 wrote:

Why are we assuming that 120 ways to distribute the remaining vouchers, after Mrs Smith has already distributed 2 vouchers per head?

Because 2 vouchers per nephew can be given in 1 way only. You distribute the others in N ways and multiply that by 1 so you will get N ways only.

Imagine a scenario in which you have a packet of 20 candies. There are 4 children to whom you want to distribute these candies in such a way that each child gets at least 2 candies. What will you do? First you will hand over 2 candies each to the 4 children. You have no choice in this.
Then you have 12 candies. Now you have a choice. You can give all 12 to your favourite child, 11 to your favourite and 1 to your second favourite and so on... All the different ways in which you distribute the candies stem from what you do with these 12 candies.

It is the same logic here. How you distribute the 7 leftover vouchers decides the number of different ways in which you can distribute the 15 vouchers.
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Mrs. Smith has been given film vouchers. Each voucher allows  [#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

given: 4 kids, at least 2 vouchers, 120 ways to distribute them;

$$k_1+k_2+k_3+k_4=vouchers… (k_1'+2)+(k_2+2)+(k_3+2)+(k_4+2)=v… k_1'+k_2+k_3+k_4=v-8$$
$$C(n+r-1,r-1)=120…C(v-8+4-1,4-1)=120…C(v-5,3)=120$$
$$\frac{(v-5)!}{3!(v-5-3)!}=120…\frac{(v-5)!}{(v-8)!}=120•3!…\frac{(v-5)(v-6)(v-7)(v-8)!}{(v-8)!}=720…(v-5)(v-6)(v-7)=720$$
now $$(v-5)(v-6)(v-7)$$ is equal to three consecutive integers, so, $$v$$ must be divisible by 3 and $$v>7$$;
test $$v=15…(10)(9)(8)=720$$ Mrs. Smith has been given film vouchers. Each voucher allows   [#permalink] 26 Aug 2019, 12:39

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