Permutation of Fermat

Probability Level 3

How many permutations of the word FERMAT have EXACTLY ONE letter in the correct place?

correct place: F (first letter), E (second letter),....


The answer is 264.

Paul Fournier by Paul Fournier , 70, Canada

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1 solution

Let D n D_{n} denote the number derangements of n objects.
The number of ways exactly one letter is in correct place is,
( 6 1 ) D 5 = 6 ⋅ 5 ! ⋅ ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + 1 4 ! − 1 5 ! ) = 264 \dbinom{6}{1} D_{5} = 6\cdot 5!\cdot \left(1-\dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} - \dfrac{1}{5!} \right) = 264

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