Permutations and combinations

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Permutations and combinations

Asked by navyaankita121993 | 08 Feb, 2021, 20:40: PM

Expert Answer

It is given that a_k does not have k element as a permutation.

It has 5 elements each with 2 possibilities (one of being the permutation and the other of not being the permutation).

So for each 5 elements 1, 2, 3, 4, 5, there are 2 possibilities.

Hence, the total number of permutations possible is 6! - ⁶C₁ x 2⁵ = 528

Since $a_{k}$ does not have $k$ element as a permutation, it has $5$ elements each with 2 possibilities (one of being in the permutation and the other of not being the permutation). Hence, total permutations possible are

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JEE Class main Answered

Since a k​ does not have k element as a permutation, it has 5 elements each with 2 possibilities (one of being in the permutation and the other of not being the permutation). Hence, total permutations possible are

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Since $a_{k}$ does not have $k$ element as a permutation, it has $5$ elements each with 2 possibilities (one of being in the permutation and the other of not being the permutation). Hence, total permutations possible are