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

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

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New post 25 Feb 2019, 04:33
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: \(x=k+8=7+8=15\).

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 \(n+r-1C_{r-1}\).

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

Hi,

let's assume that Mr. Smith has 15 vouchers and 4 nephews whom she want to give the vouchers with the restriction that everyone gets at least 2 vouchers.

No we should use the formula ("The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\)"

This would be \(15-2C_{4-1}\). This does not result in 120. Where am I wrong?
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Re: Mrs. Smith has been given film vouchers. Each voucher allows  [#permalink]

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New post 26 Feb 2019, 06:43
Debashis Roy wrote:
Bunuel VeritasKarishma
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?

Please explain


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

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

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New post 27 Feb 2019, 10:11
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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New post 27 Feb 2019, 10:15
[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
How did you arrive at 15 by solving woth that formular? Please help, I'm stuck There! Anyone hellllppp pleaseee

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

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New post 24 May 2019, 00:59
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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New post 31 May 2019, 11:16
chetan2u, VeritasKarishma, Bunuel, Gladiator59, generis

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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New post 31 May 2019, 19:46
PriyankaPalit7 wrote:
chetan2u, VeritasKarishma, Bunuel, Gladiator59, generis

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

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New post 26 Aug 2019, 13:39
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 \(v=15…(10)(9)(8)=720\)

Answer (C)
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Mrs. Smith has been given film vouchers. Each voucher allows   [#permalink] 26 Aug 2019, 13:39

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