Alg2
what is the number of possibilities of arrangements of the letters in the word intercept?

When there is no repetition of the letters, the number of permutations (order counts) is nPr, taking r letters at a time n!/(nr)!.
If all n letters are taken (as in the present case) to make the word, nPn reduces to
n!/(nn)!=n!/0!=n!/1 = n!
However, if there are repetitions of letters, we have to reduce the number of permutations by the factor r1! for each occurrence of the duplication/repetition.
Examples:
1. How many arrangements of the six letters of the word "number" are there?
6 letters, no repetitions, 6P6=6!=720.
2. How many arrangements of the eleven letters of the word "mathematics" are there?
We note repetitions of the letters m(2), a(2), t(2), so the number of arrangements
= (11P11)/(2!2!2!)
=11!/(2!2!2!)
=4989600
Post if you need more information.posted by MathMate
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