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In order to celebrate that the midterm went well, you go to your neighborhood pub. This pub has 16 different beers on tap:

  • There are 7 beers of the pilsner type.
  • There are 5 beers of the India pale ale type.
  • There are 4 beers of the German wheatbeer type.

You order 4 different beers with at least one beer of each type. What is the number of ways in which you can do this? (The order in which you order the beers does not matter.)

(a) ${16}\choose{4}$
(b) ${{7}\choose{2}} \cdot 5 \cdot 4 + 7 \cdot {{5}\choose{2}} \cdot 4 + 7 \cdot 5 \cdot {{4}\choose{2}}$
(c) ${{16}\choose{4}} - {{7}\choose{3}} - {{5}\choose{3}} - {{4}\choose{3}} $
(d) None of the above

The answer is (b).

My guess was (d) because:

  • There must be at least one beer of each type. This is the same as calculating the number of beers that aren't of each type (it's going to be 4 of one type):

    ${{16}\choose{4}} - {{7}\choose{4}} - {{5}\choose{4}} - {{4}\choose{4}} $

I realized though, that you could also choose from only 2 types of beer and that would also satisfy what I have said above, so I am wrong.

Why is (b) correct?

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  • 3
    $\begingroup$ The possible patterns are $(2,1,1), (1,2,1), (1,1,2)$ with, I hope, obvious notation. Let's count $(2,1,1)$ as an example. There are then $\binom 72$ ways to pick the two pilsners, $\binom 51=5$ ways to pick your one ale, and $\binom 41=4$ ways to pick your wheatbeer. So...$\binom 72\times 5\times 4$ ways to populate this pattern. $\endgroup$ – lulu Feb 23 '18 at 1:54
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You could choose $2$ Pilsners, $1$ Indian Ale and $1$ German beer (PPIG), thats $\binom{7}{2} \binom{5}{1} \binom{4}{1}$.

Similarly you could choose PIIG thats $\binom{7}{1} \binom{5}{2} \binom{4}{1}$.

Or PIGG thats $\binom{7}{1} \binom{5}{1} \binom{4}{2}$.

So the answer is $\color{red}{(b)}$.

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