How many permutations are there for a sequence of 8 arguments which, none of the

Question

How many permutations are there for a sequence of 8 arguments which, none of the 8 numbers are in their correction position. For example, the solved sequence is {1,2,3,4,5,6,7,8} A sequence which should not be included in the final calculation is: {1,4,6,3,8,5,7,2} But a sequence like this one should be counted: {4,1,6,3,7,5,8,2} I know for a sequence of 8 elements, there are 8! permutations (8 nPr 8), but for the question I am asking, I know that it should be much less than that. Please show work / explain how you found the answer. Thanks.

Explanation / Answer

The first position has 7 options. (1 is not anoption) The second position has 6 options left(2 is not an option, anda digit is already used) The third position would have 5 options left. 7 * 6 * 5 * 4 * 3 * 2 *1

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