To solve the problem of selecting a President, a Treasurer, and a Secretary from a committee of 11 members, we need to determine the number of ways to choose 3 positions from 11 members where the order of selection matters (since each position is unique).
The appropriate concept to use here is permutations because the order in which we select the members for each role is important.
The formula for permutations is given by:
[tex]\[ P(n, k) = \frac{n!}{(n - k)!} \][/tex]
Where:
- [tex]\( n \)[/tex] is the total number of items to choose from.
- [tex]\( k \)[/tex] is the number of items to choose.
- [tex]\( n! \)[/tex] is the factorial of [tex]\( n \)[/tex].
- [tex]\( (n - k)! \)[/tex] is the factorial of the difference between [tex]\( n \)[/tex] and [tex]\( k \)[/tex].
In this particular problem:
- [tex]\( n = 11 \)[/tex] (the total number of committee members)
- [tex]\( k = 3 \)[/tex] (the number of positions to be filled: President, Treasurer, Secretary)
Thus, the number of permutations is:
[tex]\[ P(11, 3) = \frac{11!}{(11 - 3)!} = \frac{11!}{8!} \][/tex]
Now, let's break down the calculation:
1. Calculate [tex]\( 11! \)[/tex] (11 factorial):
[tex]\[ 11! = 11 \times 10 \times 9 \times 8! \][/tex]
2. Calculate [tex]\( 8! \)[/tex] (8 factorial):
[tex]\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
As [tex]\( 8! \)[/tex] appears in both the numerator and the denominator, it cancels out, simplifying the expression to:
[tex]\[ \frac{11 \times 10 \times 9 \times 8!}{8!} = 11 \times 10 \times 9 \][/tex]
Performing the multiplication:
[tex]\[ 11 \times 10 = 110 \][/tex]
[tex]\[ 110 \times 9 = 990 \][/tex]
Thus, the number of ways to choose and arrange the President, Treasurer, and Secretary from the 11 committee members is:
[tex]\[ \boxed{990} \][/tex]
