there are 6 members on a board of directors .if they must elect a chairperson ,a secretary ,and a treasurer,how many different slates of candidates are possible.

How about you trying one of these, I have given you several examples by now.

Hint, in this problem, there are specific positions.

No,

you can pick the chairperson in 6 ways,
for each of these you can pick the secretary from the remaining 5 people, and finally the treasurer from the remaining 4 people,
so the number of ways to pick your 3 positions is
6x5x4 = 120 ways.

there are 6 members on a boars of directors. if they must elect a chairperson. a secretary, and a treasure, how many different slates of candidates are possible?

120

Well, let's see. We have 6 members and we need to fill 3 positions - chairperson, secretary, and treasurer.

For the chairperson position, we can choose any of the 6 members.
For the secretary position, we can choose any of the remaining 5 members.
For the treasurer position, we can choose any of the remaining 4 members.

So, the total number of different slates of candidates is:

6 x 5 x 4 = 120

There are 120 possible slates of candidates. That's a lot of choices! Just make sure they don't bring clown noses and whoopee cushions into the board meetings. We want a serious board, after all.

To find out the number of different slates of candidates possible for the chairperson, secretary, and treasurer positions among the 6 board members, we can use the concept of permutations.

A permutation is an arrangement of objects in a specific order. In this case, we want to select 3 positions from the 6 board members, which means we're looking for a permutation of 6 objects taken 3 at a time.

The formula to calculate permutations is: P(n, r) = n! / (n - r)!

Where:
- n is the total number of objects (board members)
- r is the number of objects to be selected (positions to be filled)

In this scenario,
n = 6 (total board members)
r = 3 (positions to be filled)

Plugging these values into the formula, we get:
P(6, 3) = 6! / (6 - 3)!
= 6! / 3!
= (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 720 / 6
= 120

Therefore, there are 120 different slates of candidates possible for the chairperson, secretary, and treasurer positions among the 6 board members.