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

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.
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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

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)

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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Thanks for your great explanation!

(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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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!

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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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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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

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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Anyway, this is a good way to solve this tough problem! Thanks

5 stars for the "stars and bars" method! The coolest trick ever!

really great trick!!

Hi Bunuel,

Why do we assume k=5?

What would happen if we assume some other value of k?

Thanks!

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

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

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

that is what i figured. thank you.

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

Hi Bunuel,
Can you help explain the formula(or share the link if it's already done) for finding the number of ways of dividing n identical items among r persons, when each one of whom receives at least one item ?
In this case, how does the concept of separator work. To me, it suggests the formula will have to be \(n+rC_{r}\). Can you help me please understand this.