There are 13 apples, 2 pen, divided to four children(A,B,C,D) Everyone has at least one thing Q:How many method?

Question

There are 13 apples, 2 pen, divided to four children(A,B,C,D) Everyone has at least one thing Q:How many method ?

Answer 1

The number of ways to give all of the apples to exactly $2$ of the kids is $\binom42\cdot\binom{12}{1}$. In each of those cases, the two pens must go to the two kids who didn't get apples. The $\binom42$ counts the ways of deciding which two kids get apples, and the $\binom{12}{1}$ counts the ways of deciding how to divvy up $11$ apples after one goes to each of the apple-getting kids. Total for case 1: $\binom42\cdot\binom{12}{1}=72$.

Now, suppose exactly 3 kids get apples. That can be done $\binom43\cdot\binom{12}{2}$ ways. For each of these ways, there are $4$ ways to hand out the pens: one goes to the non-apple kid, and the other goes to one of $4$ kids. Total for case 2: $\binom43\cdot\binom{12}{2}\cdot 4=1056$.

Suppose all four kids get apples. That can be done $\binom{12}{3}$ different ways. In each case, the pens can go to whomever, so thats $\binom53$ ways of distributing pens. Total for case 3: $\binom{12}{3}\cdot\binom53=2200$.

I'm getting a total of 3328.

Answer 2

Break into cases based on whether the pens go to the same person or whether they go to different people.

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• Pens to same person

WLOG suppose it went to first person. We have then the system $\begin{cases} x_1+x_2+x_3+x_4=13\\x_1\geq 0\\ x_2\geq 1\\ x_3\geq 1\\ x_4\geq 1\end{cases}$ which is same as system $\begin{cases} y_1+y_2+y_3+y_4=9\\ y_1\geq 0\\ y_2\geq 0\\ y_3\geq 0\\ y_4\geq 0\end{cases}$

The system has $\binom{10+4-1}{4-1}$ and there are $4$ choices for who gets the pens, so there are $4\binom{13}{3}$ outcomes in this case.

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• Pens go to different people

WLOG suppose it went to first two people. We have then the system $\begin{cases} x_1+x_2+x_3+x_4=13\\ x_1\geq 0\\ x_2\geq 0\\ x_3\geq 1\\ x_4\geq 1\end{cases}$ which can be changed in a similar way as before.

There are $\binom{11+4-1}{4-1}$ solutions to the system and $\binom{4}{2}$ ways to pick who gets the pens for a total of $\binom{4}{2}\binom{14}{3}$ outcomes in this case.

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This gives a final count of $4\binom{13}{3}+6\binom{14}{3} = 3328$, which agrees with the other answer.