Independent solution

How to solve this Combinatorial Probability question

Setup

Setup

Partition the population into four disjoint cells. Inclusion-exclusion determines the shared cell first, after which the two single-condition cells and the outside cell follow.

nHC=9+915=3n_{H\cap C}=9+9-15=3
nHC=nCH=6,n(HC)c=3515=20n_{H\setminus C}=n_{C\setminus H}=6,\qquad n_{(H\cup C)^c}=35-15=20

Model

Model

Seven specified selections come from the three condition cells, so the eighth member must come from outside their union. The cells are disjoint, allowing their combination counts to be multiplied.

N=(64)(61)(32)(201)N=\binom{6}{4}\binom{6}{1}\binom{3}{2}\binom{20}{1}

Compute

Compute

Evaluate each independent selection factor.

(64)=15,(61)=6,(32)=3,(201)=20\binom{6}{4}=15,\quad \binom{6}{1}=6,\quad \binom{3}{2}=3,\quad \binom{20}{1}=20
N=15(6)(3)(20)=5,400N=15(6)(3)(20)=5{,}400

Answer

Answer

There are 5,400 eligible groups.

5,400(C)\boxed{5{,}400\quad\text{(C)}}