There are n #>=# 11 people. Set A must have 10 people. Set B must have 11 people. A person may belong to both A and B at the same time. How many ways can this be done?

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Answer 1 · 2018-01-23T13:42:13

#((n!)^2)/((10!)(11!)(n-10)((n-11)!)^2#

We're working with combinations, the general formula of which is:

#C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"#

We have a population of #n>=11#, with #k_A=10 and k_B=11# and no limitation on having the same people in both groups. This gives then:

#C_(n,k_A)C_(n,k_B)=((n!)/((10!)(n-10)!))((n!)/((11!)(n-11)!))#

#=((n!)^2)/((10!)(11!)(n-10)((n-11)!)^2#

To test this, let's set #n=11#. By observation, we should get a result of 11. Let's see if we get it with our formula:

#((11!)^2)/((10!)(11!)(11-10)((11-11)!)^2#

#(11!)/(10!)=11 color(white)(000)color(green)root#