Let $A=\{1,2,3,4\}$ and $B=\{1,4,9,16\}$. Then the number of many-one functions $f: A \rightarrow B$ such that $1 \in f(A)$ is equal to:
- (1) 127
- (2) 151
- (3) 163
- (4) 139
Solution:
Total functions from $A \to B = 4^4 = 256$.
Number of one-one functions = $4! = 24$.
Total many-one functions = $256 – 24 = 232$.
We require many-one functions where $1 \in \text{Range}$.
It is easier to calculate the complement: Many-one functions where $1 \notin \text{Range}$.
If $1 \notin \text{Range}$, the codomain is effectively $\{4, 9, 16\}$ (3 elements).
Total functions to these 3 elements = $3^4 = 81$.
Since domain size (4) > codomain size (3), ALL these 81 functions are many-one.
So, Many-one functions where $1 \notin f(A) = 81$.
Many-one functions where $1 \in f(A) = (\text{Total Many-one}) – (\text{Many-one where } 1 \notin f(A))$
$$ = 232 – 81 = 151 $$
Ans. (2)