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At my wife's insistence, we have twelve food and water bowls for our three cats—nine circular (i.e. with circular cross-section) and three elliptical. I've just washed them all, and I'm wondering in how many ways I can put them down for the cats. Here are the rules I've (informally) developed:

  • There are two water bowls and three food bowls for the three cats.
  • The three elliptical bowls are a set; they go down together.
  • The elliptical bowls are too shallow to be used for water bowls; they're only ever used as food bowls.

Thus, when I'm putting down food and water bowls, the two water bowls are always selected from the circular bowls. The three food bowls are either all circular or all elliptical; never a mix.

I took two approaches to solving the problem and got two very different answers; I'm not sure which, if either, is correct.

Approach 1:

First, take out the two water bowls (which must be circular); there are $\binom{9}{2} = 36$ ways to do that. Then, take out the three food bowls. For these, I can choose either three of the remaining seven circular bowls ($\binom{7}{3} = 35$ possibilities) or the three elliptical bowls ($1$ possibility). Thus, the total number of combinations of water and food bowls is $36 * (35 + 1) = 1296$.

Approach 2:

For the five bowls, I can either take out five of my nine circular bowls (two for water, three for food), or two of the nine circular bowls (for water) and the three elliptical bowls (for food). The first possibility can be done in $\binom{9}{5} = 126$ ways, and the second in $\binom{9}{2} \cdot 1 = 36$ ways; thus the total number of combinations is $126 + 36 = 162$.

Which, if either, of these is correct, and why is the other (or why are both) wrong?

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    $\begingroup$ In the second approach, you forgot to multiply 126 by the ways that the 5 bowls can be distributed for food and water. So, multiply $126$ by $\begin{pmatrix}5\\2\end{pmatrix}$ and you'll get the right answer. $\endgroup$ Mar 31 '15 at 14:21
  • $\begingroup$ Aha! Excellent! Please put that in an answer and I'll accept it :-) This has been driving me so crazy I missed my exit driving to work today :-D $\endgroup$ Mar 31 '15 at 14:24
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The problem with the second approach is that when 5 circular bowls are chosen, you must choose some of them to be water dishes and the others to be food dishes. This results in exactly $\begin{pmatrix}5\\2\end{pmatrix}=10$ ways for to distribute the 5 chosen circular bowls. With this correction, the sum becomes $$ \begin{pmatrix}9\\5\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}+\begin{pmatrix}9\\2\end{pmatrix}=1296. $$

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