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Nehemiah has 10 pamphlets which he wishes to put in 13 mail boxes. In how many different ways may he do so if two mail boxes are to have two pamphlets each while the rest have a maximum of 1 pamphlet each and all pamphlets are identical?

The answer has been given as 624624

I have come up with 36036 as an answer. How would you solve this? Looking more for approach than for the final answer.

I used (13C2)(11C6) = 36036.

Delonjnaidu
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  • Your answer looks correct. It's hard to see how they could have come up with $624624$, in part because it's divisible by $13^2$. – Barry Cipra Mar 30 '17 at 17:50

1 Answers1

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The pamphlets are identical, but the mailboxes are not. Two boxes must have two pamphlets each, and the rest have at most one, hence a valid arrangement must be of the form $(n_1, \ldots, n_{13})$ where exactly:

  1. $2$ of the $n_i$ equal $2$
  2. $6$ of the $n_i$ equal $1$
  3. $5$ of the $n_i$ equal $0$,

where $n_i$ represents the number of pamphlets in mailbox $i$. This means there are $$\binom{13}{2, 6, 5} = \frac{13!}{2!6!5!} = 36036$$ such arrangements. I do not see how the answer could be $624624$; that is far too large.

In Mathematica, the input

Length[Permutations[{0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2}]]

gives the result 36036.

heropup
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