Let A = x 1 , x 2 , … , x 7 and B = y 1 , y 2 , y 3 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 = y 2 , is equal to:

Step-by-step solution

A = x 1 ,   x 2 … . . x 7 and B =   y 1 ,   y 2 ,     y 3 Let us select 3 elements from A & connect it to y 2  Number of ways = 7 C 3   The number of ways would be equal to 7 C 3   and now we have 2 elements y 1   &   y 3   in set B to be mapped from the remaining element of set A Hence, total number of function 2 4 = 16 and out of which 2 would be "into" functions (when all four goes in y 1 & when all four goes in y 3 ) = 2 4 - 2 = 14 ∴ So the