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There are 15 members in a committee. In how many different ways can a president, vice president, secretary, and treasurer be chosen?

P(n, r) = [tex]\(\frac{n!}{(n-r)!}\)[/tex]

For n = 15 and r = 4:
P(15, 4) = [tex]\(\frac{15!}{(15-4)!}\)[/tex]

Find the number of distinguishable permutations of the letters in "CAT".

Sagot :

GinnyAnsweridn
Sure, let's break down the solutions to the two parts of the problem step-by-step.

### Part 1: Number of Ways to Choose Committee Members

We are asked to find the number of ways to choose a president, a vice president, a secretary, and a treasurer from 15 members in a committee. This is a permutation problem because the order in which we choose the positions matters. The formula for permutation [tex]\( P(n, r) \)[/tex] is given by:

[tex]\[ P(n, r) = \frac{n!}{(n-r)!} \][/tex]

Where:
- [tex]\( n \)[/tex] is the total number of members,
- [tex]\( r \)[/tex] is the number of positions to fill.

In this case, [tex]\( n = 15 \)[/tex] and [tex]\( r = 4 \)[/tex]. Plugging in these values, we get:

[tex]\[ P(15, 4) = \frac{15!}{(15-4)!} = \frac{15!}{11!} \][/tex]

To simplify [tex]\( \frac{15!}{11!} \)[/tex]:

[tex]\[ \frac{15!}{11!} = \frac{15 \times 14 \times 13 \times 12 \times 11!}{11!} \][/tex]

The [tex]\( 11! \)[/tex] in the numerator and the denominator cancel each other out, leaving us with:

[tex]\[ 15 \times 14 \times 13 \times 12 \][/tex]

When we multiply these values, we get:

[tex]\[ 15 \times 14 = 210 \][/tex]
[tex]\[ 210 \times 13 = 2730 \][/tex]
[tex]\[ 2730 \times 12 = 32760 \][/tex]

So, there are 32,760 different ways to choose a president, a vice president, a secretary, and a treasurer from the 15 committee members.

### Part 2: Number of Distinguishable Permutations of "CAT"

Next, we are asked to find the number of distinguishable permutations of the letters in the word "CAT". The word "CAT" consists of 3 unique letters. To determine the number of permutations, we use the formula for permutations of distinct objects, which is just [tex]\( n! \)[/tex]. Here, [tex]\( n \)[/tex] is the number of letters.

For the word "CAT", [tex]\( n = 3 \)[/tex]:

[tex]\[ 3! = 3 \times 2 \times 1 = 6 \][/tex]

So, there are 6 distinguishable permutations of the letters in the word "CAT".

### Summary

1. The number of different ways to choose a president, a vice president, a secretary, and a treasurer from 15 committee members is 32,760.
2. The number of distinguishable permutations of the letters in "CAT" is 6.
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