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Question

Let A={x1,x2,x3,x4,x5,x6} and f:AA. The number of bijective functions such that f(xi)=xi for exactly four of the xi is

A
15.0
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B
15
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C
15.00
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Solution

Given : f(xi)=xi for exactly four elements out of six.

Way of selecting such four elements =6C4
And the rest two elements have two way of mapping i.e., (x5x5 & x6x6 or x5x6 & x6x5)


But, it is given only four elements satisfy f(xi)=xi, so one way get cancelled out.
Total number of ways =16C4=15

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