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

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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[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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Same question as last post

Using the direct nCr formula does not equate to 120 ...

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.
_________________

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.
_________________

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)

I tried to solve this in a different way but essentially ended up with the formula recommended by Bunuel.

More than 8 tickets are being distributed. if 'x' is the total number of tickets, and x-8=k (k is the remaining tickets)

So we need to distribute the k tickets to four people. Take three '|' to represent 4 people and now, we have k+3 elements from which we choose 3 '|'

(K+3)C(3) = (K+3)!/(3!*K!) = 120

(K+3)(K+2)(K+1) = 12 * 6 = 720

From answer choices

13, 14, 15, 16 --> removing the 8 from each option --> 5, 6, 7, 8 --> only 7 fits the equation..

So, if you are trying to distribute K elements across R buckets, then you need R-1 separators.
Then R-1 separators are chosen from a total of K+R-1 elements in (K+R-1)C(R-1) ways

Assume there are n identical fruits for distribution among 4 kids.
Given, each kid will receive 2 fruits (minimum, compulsorily)
Therefore, (n-8) identical fruits are open to distribution among 4 kids.

Applying the scale method:
No of scales = No of people - 1
Here, scales = 4-1 = 3

So, no of ways of distribution = (n-8+3)! / (n-8)!*3! = 120

Plug in answer choices:
A) n= 13
n-8= 5
8!/5!*3! = 56

B) n= 14
n-8= 6
9!/6!*3! = 84

C) n=15
n-8 = 7
10!/7!*3! = 120
Correct

Ans C

Went through the Stars and Bars method for Distributing Identical Items among Different Groups. Then I started from 16 and worked my way down Checking Answer Choices.


She Receives X Film Vouchers. She distributed the X IDENTICAL Film Vouchers among her 4 nephews (A , B , C , D)

A + B + C + D = X


However, each nephew must receive AT LEAST 2. So to handle this, give each nephew 2 of the Identical Items right off the bat and Reduce the No. of Identical vouchers we need to distribute by that amount.

a' + b' + c' + d' = (X - 2 - 2 - 2 - 2) or (X - 8)


Each Variable Stands in for 1 of the 4 Different Nephew "Groups" we are distributing to.
Each will actually receive:
A = a' + 2
B = b' + 2
C = c' + 2
D = d' + 2


To find the No. of Ways to Distribute the Identical Items, find the No. of Ways you can Arrange the (X - 8) IDENTICAL Voucher Elements and the 3 "Plus Signs / Vertical Separators".

The total No. of Elements to be arranged = (X - 8) Identical Vouchers + 3 Identical Vertical Seperators

Since these Elements are Indistinguishable, we must Divide by the No. of Identical Elements to remove the Arrangements that are IDENTICAL.

Total No. of Ways to Distribute = 120 = (X - 8 + 3)! / 3! * (X - 8)!


-E- if she received 16 Identical Vouchers

Does: 120 = (16 - 8 + 3)! / 3! * (16 - 8)!

120 = (11)! / 3! * 8!

120 = 11 * 10 * 9 / 3 * 2 * 1

120 does NOT = 11 * 5 * 3




-D- 15

Does: 120 = (15 - 8 + 3)! / 3! * (15 - 8)!

120 = 10! / 3! * 7!
120 = 10 * 9 * 8 / 3 * 2
120 = 10 * 3 * 4
120 = 10 * 12

Answer C - 15

Has this type of question been seen on gmat before? seems a bit difficult and time consuming, given that even if you know the formula, if you dont catch the "consecutive numbers as factors" quickly, youll be stuck solving a fairly complex problem