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Examples: 1. Find the number of permutations of the letters of the word STATISTICS. Solution: There are 10 letters of the word. Assuming that the letters are distinct, there are P(10,10) = 10! permutations. However, we have to take into considerations that the 3 S's are alike, the 3 T's are alike, and the 2 I's are also alike. The permutations of the 3 S's is P(3,3) = 3!. The permutations of the 3 T's is P(3,3) = 3! The permutations of the I's is P(2,2) = 2!. So we must divide 10! by 3! 3! 2! in order to eliminate the duplicates. Thus, $P = \frac{10!}{3!3!2!}$

          Examples:
1. Find the number of permutations of the letters of the word STATISTICS.
Solution:
There are 10 letters of the word. Assuming that the letters are distinct, there are
P(10,10) = 10! permutations.
However, we have to take into considerations that the 3 S's are alike, the 3 T's are
alike, and the 2 I's are also alike. The permutations of the 3 S's is P(3,3) = 3!. The
permutations of the 3 T's is P(3,3) = 3! The permutations of the I's is P(2,2) = 2!.
So we must divide 10! by 3! 3! 2! in order to eliminate the duplicates. Thus,
$P = \frac{10!}{3!3!2!}$

Examples:
1. Find the number of permutations of the letters of the word STATISTICS.
Solution:
There are 10 letters of the word. Assuming that the letters are distinct, there are
P(10,10) = 10! permutations.
However, we have to take into considerations that the 3 S's are alike, the 3 T's are
alike, and the 2 I's are also alike. The permutations of the 3 S's is P(3,3) = 3!. The
permutations of the 3 T's is P(3,3) = 3! The permutations of the I's is P(2,2) = 2!.
So we must divide 10! by 3! 3! 2! in order to eliminate the duplicates. Thus,
P = (10!)/(3!3!2!)

Added by Danny H.

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