Let A={x1, x2, ..., x7} and B={y1, y2, y3} be two sets containing seven and three distinct elements respectively. Then the total number of functions f: A→B that are onto. If there exist exactly three elements x in A such that f(x) = y2, is equal to:

Given, there are exactly three elements in A such that f(x) = y2.
Number of ways of selecting 3 elements from A = 7C3
Now the each remaining element in A can be associated with either of y1, y3 in B.
Total No of ways = 2 × 2 × 2 × 2, but subtracting the cases where all the remaining elements getting associated to either y1 or y3, we have 16 − 2 = 14 ways.
Therefore total number of onto functions = 14 × 7C3