ways of choosing 3 elements of set x ( for f(x) =b1) = 6C3 = 20
now we are left with x={A,B,C} and y={M,N}
number of onto functions in this case = 2*3C2*1 = 6
thus total number of functions = 20*6 =120
Let x={a1,a2,a3,......,a6} and y={b1,b2,b3}. Then number of functions f from x to y such that it is onto and there are exactly three elements in x such that f(x) =b1, is
a) 75 b) 90
c) 100 d) 120
Ans: (d)
Please explain how to get answer?
I am not getting by formula.
ways of choosing 3 elements of set x ( for f(x) =b1) = 6C3 = 20
now we are left with x={A,B,C} and y={M,N}
number of onto functions in this case = 2*3C2*1 = 6
thus total number of functions = 20*6 =120