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.

  • 1
    $\begingroup$ 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! $\endgroup$ – André Nicolas Aug 16 '13 at 6:24
  • $\begingroup$ 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. $\endgroup$ – Kasper-34 Aug 16 '13 at 6:30
  • $\begingroup$ 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 $5!$. $\endgroup$ – André Nicolas Aug 16 '13 at 6:33
  • $\begingroup$ 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 $5,3,1,2,4$? $\endgroup$ – Kasper-34 Aug 16 '13 at 6:41
  • 1
    $\begingroup$ Okay, in that case I agree with you. The use of the word "respectively" makes me think they must be in the order $1,2,3,4,5$ and only that order. $\endgroup$ – Kasper-34 Aug 16 '13 at 7:10

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.

  • $\begingroup$ 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 $5!$. $\endgroup$ – Kasper-34 Aug 16 '13 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.