VThe Mathematics Club will select a president, a vice president, and a treasurer for the club. If there are 15 members in the club, how many different selections of a president, a vice president, and a treasurer are possible if each club member can be selected to only one position?

2730 different selections

Step-by-step explanation:

This problem is solved using permutations: it is similar to combination, but the order of each element matters (if person A is president, person B is vice and person C is treasurer, this is a different case from a case where person A is vice, person B is treasurer and person C is president)

The formula of permutation is:

P = n!/(n-p)!

where n is the total number of members in this case (15), and p is the number of different positions (3).

So, the number of different selections is:

P = 15!/12! = 15*14*13 = 2730 different selections

50

step-by-step explanation:

only 10/60 isn't equivalent

hope this : )