# 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?

• It would help people give you good answers if you could explain why you think they should be equal – postmortes Jul 25 at 8:32
• @postmortes I have edited my question, hope it's much clearer! – Hannah Tang Jul 25 at 8:37
• Given your revised question, you should change your title. Do you wish to ask why $\binom{20}{12} \neq \binom{20}{11}\binom{9}{1}$ or why the number of ways to pick a jury differs from the number of ways of choosing a jury with a foreman? – N. F. Taussig Jul 25 at 8:41
• Do you mean $$\binom{6}{3}=\binom{6}{2}\binom{4}{1}$$? – Dr. Sonnhard Graubner Jul 25 at 8:41
• @Dr.SonnhardGraubner That is what she means. However, she has revised her question since you first read it. – N. F. Taussig Jul 25 at 8:42

One can also verify that: $$12\times\binom{20}{12}=\binom{20}{11}\binom{9}{1}$$
• Your equation should be $12\times \binom{20}{12}=\binom{20}{11}\binom{9}{1}$ rather than $\binom{20}{12}=12\times\binom{20}{11}\binom{9}{1}$ as you want to designate one of the 12 jury members as foreman. – parafoo Jul 25 at 9:13