Question

Let $A=\{a_1,a_2,a_3,a_4{a}_5,a_6\}$ and $B=\{b_1,b_2,b_3\}$. The number of functions of $f:A\to B$ such that it is onto and there are exactly three elements in $A$ such that $f\left(A\right)=b$, is

• A. $75$
• B. $90$
• C. $100$
• D. $120$

Answer 1

Answer: (D)

$120$

Since there should be exactly 3 elements in $A$ such that

$f\left(A\right)=b$

the function $f$ is a bijection from a subset $A_1$ of $A$ to $B$., each having 3 elements.

Number of ways to select 3 elements from $A$ $={}^6{C}_3$

Number of bijection from $A_1$ to $B$ $=3!$

Hence, the number of functions $={}^6{C}_{3}3!$ $=120$