To determine how many different four-letter permutations can be formed using four letters out of the first twelve in the alphabet, we can use the formula for permutations. The formula for permutations is given by:
[tex]\[
{}_nP_r = \frac{n!}{(n-r)!}
\][/tex]
where:
- [tex]\( n \)[/tex] is the total number of items to choose from,
- [tex]\( r \)[/tex] is the number of items to choose,
- [tex]\( n! \)[/tex] (n factorial) is the product of all positive integers up to [tex]\( n \)[/tex],
- and [tex]\( (n-r)! \)[/tex] ((n-r) factorial) is the product of all positive integers up to [tex]\( n-r \)[/tex].
In this problem:
- [tex]\( n = 12 \)[/tex] (since we are choosing from the first twelve letters in the alphabet),
- [tex]\( r = 4 \)[/tex] (since we need to form four-letter permutations).
Using the permutation formula:
[tex]\[
{}_nP_r = \frac{12!}{(12-4)!} = \frac{12!}{8!}
\][/tex]
To solve this, we need to calculate [tex]\( 12! \)[/tex] and [tex]\( 8! \)[/tex] and then divide the two results.
[tex]\[
12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
[tex]\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
Since [tex]\( 8! \)[/tex] is a subset of [tex]\( 12! \)[/tex], it simplifies our calculation. We can cancel out [tex]\( 8! \)[/tex] from both numerator and denominator:
[tex]\[
\frac{12!}{8!} = 12 \times 11 \times 10 \times 9
\][/tex]
Multiplying these values together, we get:
[tex]\[
12 \times 11 = 132
\][/tex]
[tex]\[
132 \times 10 = 1320
\][/tex]
[tex]\[
1320 \times 9 = 11880
\][/tex]
So, the number of different four-letter permutations that can be formed using four letters out of the first twelve in the alphabet is:
[tex]\[
11880
\][/tex]
