Let A = x 1 , x 2 , … , x 7   and   B = y 1 , y 2 , y 3 be two sets containing seven and…

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:AB that are onto, if there exist exactly three elements x in A such that fx=y2, is equal to:
  1. 12·7C2
  2. 16·7C3
  3. 14· 7C3
  4. 14·7C2

Solution

A={x1, x2..x7} and B= {y1, y2,  y3}

Let us select 3 elements from A & connect it to y

Number of ways =7C3 

The number of ways would be equal to 7C3  and now we have 2 elements y1 & y3  in set B to be mapped from the remaining element of set A

Hence, total number of function 24=16 and out of which 2 would be "into" functions (when all four goes in y1 & when all four goes in y3) =24-2=14

So the total number of onto functions would be 14·7C3  

Asked in: JEE Main 2015 (11 Apr Online)

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