JEE MATHEMATICS MCQ: PERMUTATIONS

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Application +4 | -1

Let

A = \left\{ {{x_1},{x_{2,}}{x_3},\;{x_4},\;{x_5},\;{x_6}} \right\}

B = \left\{ {{y_1},\;{y_2},\;{y_3},{y_4},{y_5},{y_6}} \right\}.

. Then the number of one – one mapping from

A

B

such that

f\left( {{x_i}} \right) \ne {y_v}i = 1,\;2,\;3,\;4,\;5,\;6

View answer and explanation

Correct option: B

265

The correct answer is 265 because it is calculated by using the principle of inclusion-exclusion to find the number of one-to-one mappings where no element maps to itself.

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