Prentice has five daughters and ten identical pens. In how many ways can the pens be distributed among his daughters if two of them, Charlotte and Emily, must get the same number of pens, and every daughter is not required to get a pen?

Solutions

Charlotte(C) and Emily(E) have the same number of pens and possible numbers of pens they each can get is 0, 1, 2, 3, 4, or 5. Let us first consider the case that C and E get 0 pens. Other cases are similar.

C and E do not get pens

In this case, C and E do not get pens (i.e., each gets 0 pens). Ten remaining pens are distributed among three other daughters. How many ways are there?

This is a variation of the stars and bars problem. Since a daughter does not have to get a pen (0 is allowed), according to Theorem two on the page, the number of ways to distribute \(n\) indistinguishable pens among \(d\) daughters is

\[{n + d - 1 \choose d - 1}.\]

So there are \({ 10 + 3 - 1 \choose 3 - 1} = { 12 \choose 2 } = 66\) ways.

List all cases

Now we can consider all cases. This is summarized in the table.

Number of pens each C and E gets Number of ways distributing remaining pens
0 \({ 10 + 3 - 1 \choose 3 - 1} = { 12 \choose 2 } = \frac{12 \times 11}{2} = 66\)
1 \({ 8 + 3 - 1 \choose 3 - 1} = { 10 \choose 2 } = \frac{10 \times 9}{2} = 45\)
2 \({ 6 + 3 - 1 \choose 3 - 1} = { 8 \choose 2 } = \frac{ 8 \times 7}{2} = 28\)
3 \({ 4 + 3 - 1 \choose 3 - 1} = { 6 \choose 2 } = \frac{ 6 \times 5}{2} = 15\)
4 \({ 2 + 3 - 1 \choose 3 - 1} = { 4 \choose 2 } = \frac{ 4 \times 3}{2} = 6\)
5 \({ 0 + 3 - 1 \choose 3 - 1} = { 2 \choose 2 } = \frac{ 2 \times 1}{2} = 1\)

The answer is 66 + 45 + 28 + 15 + 6 + 1 = 161.

Discussions

It seems you need to know the stars and bars problem to solve this problem.

If everyone has to get at least one pen, use Theorem one, instead of Theroem two, on the page of stars and bars problem.