Probability of Forming a committee of 4 from nine men to always include a particular man. My Attempt: 9 nCr3. Would the fundamental counting theorem be used in this problem as one might consider that the particular man is being paired with the other 8 people? 
 A: To count the favourable ways, take the particular man out. Now choose $3$ from the remaining $8$, in $$\dbinom{8}{3}$$ ways. Since the possible ways are $\dbinom{9}{4}$ you have that the required probability is equal to $$\dfrac{\dbinom{8}{3}}{\dbinom{9}{4}}=\frac{8!4!5!}{3!5!9!}=\frac{4}{9}$$
A: Let the 9 candidates be $\{1,2,3,\ldots,9\}$.
By symmetry, every candidate has the same probability to be in the committee. Let's call that probability $p$. (This is the expectation of a random variable whose value is always either 0 or 1, depending on whether Mr. 1 is on the committee or not. The expectation will be somewhere between those two possible values).
Since candidate $1$ has probability $p$ to be in the committee, the expectation of the number of elements of the set $\{1\}$ that are on the committee is $p$.
Likewise the expectation of the number of committee members in $\{2\}$ is $p$. And so forth.
The total number of members of the committee is (the number of committee members in $\{1\}$) plus (the number of committee members in $\{2\}$) plus ... up to (the number of committtee members in $\{9\}$).
Since expectations are additive, the expected number of members of the committee is $p+\cdots+p=9p$. But we know we're always selecting exactly 4 members, so $9p=4$. Therefore $p$ must be $4/9$.
(Look, no binomial coefficients!)
