Combinatorics: Why isn't $(^{20}C_{12}) ≠ (^{20}C_{11})(^9C_1)$? I came across $2$ questions: 1. A jury pool of $20$ people are called to a courthouse. How many ways are there to select $12$ to serve as jury? 2. A jury pool of $20$ people are called to a courthouse. How many ways are there to select $11$ to serve as jury, and $1$ to serve as jury foreman? Can someone give me an intuitive explanation on how to think about this? Mathematically I know the calculation returns different values but I don't really understand why. It can't be because of the distinction between "jury" and "jury foreman", right? Ultimately we're still selecting $12$ people from a pool of $20$ isn't it?

The difference lies in making the decision on who is the jury foreman. In the first case, all the 12 selected people are equal. In the second, one of the selected 12 needs to be designated the jury foreman, which can be then done in 12 different ways. So, the answer to the second should be 12 times the answer to the first.

One can also verify that: $$12\times\binom{20}{12}=\binom{20}{11}\binom{9}{1}$$