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Question

Let A={x1,x2,x3,....x7}, B={y1,y2,y3.} The total number of functions f:AB such that every element in B has atleast one pre image and there are exactly three element x in A such that f(x)=y2 is equal to

A
490
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B
510
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C
630
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D
none of these
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Solution

The correct option is A 490
Three elements from set 'A' can be selected in 7C3 ways. Their image has to be y2. Remaining 2 images can be assigned to remaining 4 pre-images in 24 ways. But set B has atleast one pre image for every element of it, hence the number of ways is 242. Then the total number of functions is 7C3×14=490.

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