What is the probability that the owner is on the review committee and there is at least one member from each of the other three staff categories? 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 the owner is on the review committee and there is at least one member from each of the other three staff categories on the committee too? Their answer: $\frac{68439}{22\choose8}$ The only way I can think of doing it would take too long. Say you put one in each but then add all the options where the other 4 are chosen from different staff categories. That means I would need to add many different options. It's not impossible since it's only 4 people but surely there's a more efficient method?

To get a desirable outcome, we must select the owner for the committee. That leaves 7 committee members left to select from the remaining 21 employees. Let $A, B, C$ be the events that there are no Senior, no Junior, and no Support personnel (respectively) on the committee. So the complement of having all groups represented is $A\cup B\cup C$. In what follows I use $n(S)$ for the number of elements in a set $S$. Using the Principle in Inclusion/Exclusion, we have that $$\begin{array}{ccl}n(A\cup B\cup C)&=&n(A)+n(B)+n(C)-n(A\cap B) - n(A\cap C) - n(B\cap C) + n(A\cap B\cap C)\\\ &=& \begin{pmatrix} 18\\\7 \end{pmatrix} +\begin{pmatrix} 7\\\7 \end{pmatrix}+\begin{pmatrix} 17\\\7 \end{pmatrix}-\begin{pmatrix} 14\\\7 \end{pmatrix}\\\ &=& 47841 \end{array}$$

This is the complement of what we want. So the number of ways to select a desireable committee is $\begin{pmatrix} 21\\\7 \end{pmatrix}-47841=68439$.