# Distinguishable objects into distinguishable boxes

How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them respectively?

$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ \\ &1 &2 &3 &4 &5 \end{gather}$

The lines represent the $5$ distinguishable boxes and the numbers below represent how many distinguishable objects each box must hold. I'm thinking I have $C\left(15,1\right)$ options for the first box then $C\left(14,2\right)$ for the second box, all the way to $C\left(5,5\right)$ for the fifth box. I multiply all those combinations together because of the product rule and I have no idea if that's the right answer.

• Good clear correct analysis. I would say that for each option for the first box there are $\dots$. Surely it is not true that you have no idea whether this is the right answer! – André Nicolas Aug 16 '13 at 6:24
• I don't see where you're going with that ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have reached the correct answer. – Kasper-34 Aug 16 '13 at 6:30
• I just meant it should be made clearer why we multiply. Note that if it is not specified which boxes contain $1,2,\dots,5$ then we need to multiply your answer by $5!$. – André Nicolas Aug 16 '13 at 6:33
• Well I'm multiplying because of the product rule. I think? I would multiply by $5!$ if I wasn't restricted, because I could put them in any order such as $5,3,1,2,4$? – Kasper-34 Aug 16 '13 at 6:41
• You know you are correct. :) – Alex Strife Aug 16 '13 at 6:46

Ways to put the labels $\{1,2,3,4,5\}$ on the boxes according as how many objects they contain: 5!. Then, as you correctly presumed,

$\binom{15}{1}$ ways to select an object for the one-object box;

$\binom{14}{2}$ ways to select two objects for the two-object box;

$\binom{12}{3}$ ways to select three objects for the three-object box;

$\binom{9}{4}$ ways to select four objects for the four-object box;

$\binom{5}{5}=1$ way to put the remaining five objects into the five-object box.

I think the answer is $$5!\binom{15}{1}\binom{14}{2}\binom{12}{3}\binom{9}{4}.$$

If the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the term $5!$ should be suppressed.

• I would say in this particular case since it used the word "respectively" there is only one way to order the boxes, which means we can leave out the multiplication of $5!$. – Kasper-34 Aug 16 '13 at 7:17