A president, treasurer, and secretary (all different) are to be chosen from a club consisting of 10 people. How many different choices of officers are possible if:
i. there are no restrictions
- = $3!{10 \choose 1,1,1}$
3 arrangements x choosing 3 out of 10 people
ii. A and B will not serve together?
- = $3![{8 \choose 3}{1 \choose 0}{1 \choose 0}+{8 \choose 2}{1 \choose 1}{1 \choose 0}+{8 \choose 2}{1 \choose 0}{1 \choose 1}]$
choosing neither A nor B + choosing A (not B) + choosing B) x 3 arrangements
iii. C and D will serve together or not at all?
- = $3![{8 \choose 3}{1 \choose 0}{1 \choose 0}+{8 \choose 1}{1 \choose 1}{1 \choose 1}$
(choosing neither C nor D + choosing C and D and 1 other) x 3 arrangements
iv. E must be an officer?
- = $3!{9 \choose 2}$
v. F will serve only if he is president?
- = $3![{9 \choose 3}{1 \choose 0}]+2[{9 \choose 2}{1 \choose 1}]!$
Does my work look correct?
Best Answer
$\color{green}{\checkmark}$ Well, other than the first, all solutions look correct.
The first should have been $\,3!\binom{10}{3}\,$, or $\,3!\;{^{10}C_3}\,$, which is permutations of selecting 3 of 10 people.
Alternatively, this is the multinomial $\;\binom{10}{1,1,1,7} \;= \frac{10!}{7!}\;$, to count ways to select one person in each subsequent position (and 7 remainders) out of 10 people.
Addendum: Also the bang symbol in the last answer is seriously in the wrong place. I suspect a typo.