Question

a committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions.​

Answer 1

Answer:

$\frac{1}{990}$

Step-by-step explanation:

The full question:

"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current​ positions?"

The permutation of choosing 3 members from a group of 11 would be:

P(n,r) = $\frac{n!}{(n-r)!}$

Where n would be the total [in this case n is 11] & r would be 3

Which is:

P(11,3) = $\frac{11!}{(11-3)!}=\frac{11!}{8!}=11*10*9=990$

So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:

1/990