# Alg2

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

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3. 6
1. When there is no repetition of the letters, the number of permutations (order counts) is nPr, taking r letters at a time n!/(n-r)!.
If all n letters are taken (as in the present case) to make the word, nPn reduces to
n!/(n-n)!=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

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posted by MathMate

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