How many ways can 6 committee members claim 12 donuts?

Question

A box of 12 identical donuts is brought to a meeting with six committee members. How many ways can the six members claim the donuts? (A committee member can claim 0 donuts, and not all the donuts have to be claimed.)

Answer 1

Add a virtual seventh member which eats all remaining donuts. Therefore we have to allocate $12$ identical donuts to $7$ people.This is a standard problem. Now find out what kind, and you'll know the solution.

Answer 2

Let $x_i$ and $1\le i\le 6$ be the number of donuts claimed by the six people of the committee. As all the donuts need not to be claimed we introduce new variable $x_7$ which takes on the extra donuts left out after the six members have claimed their donuts.

Hence by star and bars method we need to find the non negative integral solutions of the equation

$$\sum_{i=1}^7 x_i=12$$

By using star and bars we get the answer as $$\binom {18}{6}=18564$$

Answer 3

I intend to use the multinomial theorem to solve this problem.

First let's concoct the multinomial expression. To do that we need to understand the core of the problem.

• There are six committee members, and it is not necessary that all donuts have to be claimed.

• A committee member may not claim a donut and cannot claim more than 12 donuts.

Therefore the multinomial expression as a consequence is,

$(1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6} + ... \infty)^{6} $

Now, for the sake of simplicity we will find the co-efficient of $x^{n}$ in the above expression.

The above expression is an infinite geometric series which can also be written as :

$(1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6} + ... \infty)^{6} = $ $\dfrac{1}{(1-x)^{6}} = $ $(1-x)^{-6}$

We can express $(1-x)^{-6}$ in Taylor Series as :

$(1+ {6 \choose 1}x + {7 \choose 2}x^{2}+ {8 \choose 3}x^{3} + .... + {n+5 \choose n}x^{n} + ... \infty) $

Therefore the co-efficient of $x^{n}$ is ${n+5\choose n}$.

Now the number of ways in which a total of $n$ donuts are claimed by $6$ committee members is ${n+5\choose n} = {n+5\choose 5}$

Now we need to sum up all the cases from $n=0$ to $n=12$, which is given by,

$$\sum_{n=0}^{12} {n+5\choose 5} = 18564 $$

Therefore the total number of ways by which the members of the committee can claim the 12 donuts is $18564$.