Q. 1.3 - A First Course in Probability

Question

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

(a) there are no restrictions?

(b) A and B will not serve together?

(c) C and D will serve together or not at all?

(d) E must be an officer?

(e) F will serve only if he is president?

Step-by-Step Solution

Answer

(a) The different choices of officers are possible if there are no restrictions is $720$.

(b) The different choices of officers are possible if A and B will not serve together is $672$.

(c) The different choices of officers are possible if C and D will serve together or not at all is $384$.

(d) The different choices of officers are possible if E must be an officer is $216$.

(e) The different choices of officers are possible if F will serve only if he is president is $576$.

Part (a) Step 1. Find possible no. of different choices of officers if there are no restrictions.

A president, treasurer, and secretary, all different, are to be chosen from a club consisting of $10$ people.

We have to find possible no. of different choices of officers  if there are no restrictions.

$3$ people out of $10$ can be chosen in $\left(\begin{array}{c}10\\ 3\end{array}\right)\text{ways} = \frac{10!}{3!7!}\text{ways}$

The three chosen people can be arranged in $3$ job titles in $=3! \text{ways}$

Therefore, the possible no. of choices are $=3!\times \frac{10!}{3!7!}=3!\times \frac{10\times 9\times 8\times 7!}{3!7!}=720$

Part (b) Step 1. Find possible no. of different choices of officers if A and B will not serve together.

Total no. of persons $=10$

If we exclude A and B, then total no. of persons $=8$

Out of $8$, $3$ members can be selected in $\left(\begin{array}{c}8\\ 3\end{array}\right)\text{ways}=\frac{8!}{3!5!}$

The three chosen people can be arranged in $3$ job titles in $=3!\text{ways}$

Therefore, the possible no. of choices are $3!\times \frac{8!}{3!5!}=8\times 7\times 6=336$

From A and B, either will be selected in $=\left(\begin{array}{c}2\\ 1\end{array}\right)\text{ways}=\frac{2!}{1!1!}=2$

Out of the $3$ positions, A or B can fill the positions in $\left(\begin{array}{c}3\\ 1\end{array}\right)\text{ways}=\frac{3!}{1!2!}=3$

Rest two positions can be filled in $\left(\begin{array}{c}8\\ 2\end{array}\right)\times 2! \text{ways}=\frac{8!}{2!6!}\times 2!=56$

No. of selections in which A and B will not serve together $=2\times 3\times 56=336$

Therefore, the possible no. of choices are $=336+336=672$

Part (c) Step 1. Find possible no. of different choices of officers if C and D will serve together or not at all.

The possible no. of choices if C and D are excluded $=8\times 7\times 6=336$

No. of ways in which C can fill any of the three job positions $=3$

No. of ways in which D can fill any of the two job positions $=2$

No. of ways in which the third job position can be filled $=8$

Therefore, the possible no. of choices are $=336+\left(3\times 2\times 8\right)=336+48=384$

Part (c) Step 1. Find possible no. of different choices of officers if E must be an officer .

No. of ways in which E can fill any of the three job positions $\left(\begin{array}{c}3\\ 1\end{array}\right)\text{ways}=3$

No. of ways in which remaining two job positions can be filled $\left(\begin{array}{c}9\\ 2\end{array}\right)\text{ways}=\frac{9!}{2!7!}\times 2!=9\times 8=72$

Therefore, the possible no. of choices are $=72\times 3=216$

Part (e) Step 1. Find possible no. of different choices of officers if F will serve only if he is president .

If F is President then remaining two positions can be filled in $\frac{9!}{2!7!}\times 2! \text{ways}=72$

If F is not a President then the three positions can be filled in 

$\frac{9!}{3!6!}\times 3! \text{ways}=504$

Therefore, the possible no. of choices are $=72+504=576$