# The number of ways in which 10 identical apples can be distributed to six children so that each child receives at least one apple

The number of ways in which 10 identical apples can be distributed among 6 children so that each child receives at least one apple is?

My Attempt:

I got the number possibilities to be: $(1,1,1,1,1,5), (1,1,1,1,2,4), (1,1,2,2,2,2), (1,1,1,2,2,3)$

Since they are identical apples, the number of ways each of these possibilities can be formed is 1.

However, I don't know how am I going to use this to solve the problem?

Any help would be appreciated.

• You can first give everyone an apple. Then the question becomes: "What is the number of ways you can distribute 4 apples among 6 children?" - and for this you can use the "stars and bars" method. Commented Mar 12, 2018 at 10:34
• @MattiP. The stars and bars method can be used right away. See theorem 1 here: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) Commented Mar 12, 2018 at 10:35
• You overlooked $(1,1,1,1,3,3)$. Commented Mar 12, 2018 at 10:45
• @BarryCipra Ahhh! Thanks for pointing it out. Commented Mar 12, 2018 at 10:49

There is a better approach. Let $x_k$ be the number of apples received by the $k$th child. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 10$$ is an equation in the positive integers. A particular solution corresponds to the placement of five addition signs in the nine spaces between successive ones in a row of ten ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance, $$1 1 + 1 + 1 1 1 + 1 + 1 + 1 1$$ corresponds to the solution $x_1 = 2$, $x_2 = 1$, $x_3 = 3$, $x_4 = x_5 = 1$, $x_6 = 2$. The number of such solutions is the number of ways we can select five of the nine spaces in which to place an addition sign, which is $$\binom{9}{5}$$

Addendum: Using Barry Cipra's observation, we can confirm this result by using your method.

One child receives five apples and the other five children each receive one apple: There are $6$ ways to select the child who receives five apples.

One child receives four apples, another child receives two apples, and each of the other four children each receive one apple: There are six ways to choose the child who receives four apples and five ways to choose the child who receives two apples. Hence, there are $6 \cdot 5 = 30$ such distributions.

Two children each receive three apples and the other four children each receive one apple: There are $$\binom{6}{2} = 15$$ ways to select the two children who each receive two apples.

One child receives three apples, two children each receive two apples, and the other three children each receive one apple: There are six ways to choose the child who receives three apples and $\binom{5}{2}$ ways to choose which two of the other five children each receive two apples. Hence, there are $$\binom{6}{1}\binom{5}{2} = 6 \cdot 10 = 60$$ such distributions.

Four children each receive two apples and the other two children receive one apple: There are $$\binom{6}{4} = 15$$ ways to select which two children will receive two apples.

Observe that $$6 + 30 + 15 + 60 + 15 = \binom{9}{5}$$

• Yes, I know this method and even got the answer from it. But can you please help me, how do I proceed with the given condition in the question? Commented Mar 12, 2018 at 10:38
• And strangely, the answer given in my book is $3*\binom{9}{5}$ Commented Mar 12, 2018 at 10:40
• The answer in the book is wrong. I did use the given condition. If there were no such restriction, we would have to solve the equation $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 10 \tag{1}$$ in the nonnegative integers. Commented Mar 12, 2018 at 10:48
• A particular solution of equation 1 in the nonnegative integers corresponds to the placement of five addition signs in a row of $10$ ones. For instance, $$1 1 + + 1 1 1 + 1 1 + 1 + 1 1$$ corresponds to the solution $x_1 = 2$, $x_2 = 0$, $x_3 = 3$, $x_4 = 2$, $x_5 = 1$, $x_6 = 2$. The number of such solutions is $$\binom{10 + 5}{5} = \binom{15}{5}$$ since we must select which $5$ of the $15$ positions required for $10$ ones and $5$ addition signs will be filled with addition signs. Commented Mar 12, 2018 at 10:49
• I have edited your answer to compare your approach with mine. Commented Mar 12, 2018 at 15:10