What is the probability that there are strictly fewer support staff on the committee than senior executives? A company has twenty two staff members: the owner, three senior executives, fourteen junior executives and four support staff. A salary review committee consisting of eight members of staff needs to be convened. The members of the committee are selected at random. What is the probability that there are strictly fewer support staff on the committee than senior executives. This was my approach: Where senior executives = 3, support staff <= 2 Where senor executives = 2, support staff <= 1 Where senior executives = 1, support staff = 0 Senior executives cannot equal 0 because support staff cannot be less than 0. => Is this assumption correct? P = $\frac{{3\choose3}{17\choose5}{+}{3\choose2}{16\choose6}{+}{3\choose1}{15\choose7}}{22\choose8} = \frac{49517}{22\choose8}$ Their answer was $\frac{81549}{22\choose8}$?

When senior executives = $i$, support staff = $j$, we get ${3\choose i}{4\choose j}{15\choose 8-i-j}$.

So the answer is $\dfrac{\sum_{i=1}^{3}\sum_{j=0}^{i-1}{3\choose i}{4\choose j}{15\choose 8-i-j}}{{22 \choose 8}}$

When you write ${3\choose 3}{17\choose 5}$ I think you are excluding 2 support staff and choosing 5 people among the others. But you also need to choose who are the 2 excluded support stuff, so something is missing here.