Let A=x1,x2,x3,…,x7, B=y1,y2,y3. The total number of functions f:A→B that are on to and there are exactly three element x in A such that f (x) = y2 is equal to

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.