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# Let $\mathrm{A}=\left\{{\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3},\dots ,{\mathrm{x}}_{7}\right\}$, $\mathrm{B}=\left\{{\mathrm{y}}_{1},{\mathrm{y}}_{2},{\mathrm{y}}_{3}\right\}$. The total number of functions $\mathrm{f}:\mathrm{A}\to \mathrm{B}$ that are on to and there are exactly three element x in A such that f (x) = y2 is equal to

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a
490
b
510
c
630
d
none of these

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detailed solution

Correct option is A

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 the function is onto, hence the number of ways is 24 - 2. Then the total number of functions is  7C3×14=490.

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