In a club 15 members choose from them a president , secretary and a treasurer. How many different states of candidates can be chosen?
Since these are special positions, the order counts, so
what is 15x14x13 ?
What is 15!/3!12!?
answer 15*14*13/3*2=5*7*13
Ok, I agree with Reiny, I wasn't counting them as special positions.
To determine the number of different combinations for choosing a president, a secretary, and a treasurer from a group of 15 members, we can use the concept of permutations.
The number of permutations of n items taken r at a time can be calculated using the formula:
P(n, r) = n! / (n - r)!
Where "!" denotes factorial, which means multiplying a number by all the positive integers less than it down to 1.
In this case, we need to select 3 members (president, secretary, and treasurer) from a group of 15 members, so we can calculate as:
P(15, 3) = 15! / (15 - 3)!
= 15! / 12!
Calculating the factorials:
15! = 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
By canceling out common terms, we can simplify:
P(15, 3) = (15 x 14 x 13 x 12!) / (12!)
The (12!) terms in the numerator and denominator cancel out, leaving us with:
P(15, 3) = 15 x 14 x 13 = 2,730
Therefore, there are 2,730 different states of candidates that can be chosen for the positions of president, secretary, and treasurer from the group of 15 members.