Question

Let F denote the set of all onto functions from $A=\{a_1,a_2,a_3,a_4\}$ to $B=\{x,y,z\}$. A function $f$ is chosen at random from F. The probability that $f$ is defined such that $x$ is only mapped once.

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $0$

Answer 1

Answer: (A)

$\frac{2}{3}$

Let us first count the number of elements in F.
Total number of functions from A to B is $3^4=81$.
The number of functions which contain exactly two elements in the range is $3\cdot 2^4=48$.
The number of functions which contain exactly one elements in its range is $3\times 1^4=3$ .
Thus, the number of onto functions from A to B is $81-48+3=36$ [using principle of inclusion exclusion]
$\therefore n\left(F\right)=36$
Let $f\in F$.
We now count the number of ways in which $f^{-1}\left(x\right)$ consists of single element.
We can choose preimage of $x$ in $4$ ways.
The remaining 3 elements can be mapped onto $[y,z]$ is $2^3-2=6$ ways.
$\therefore f^{-1}\left(x\right)$ will consists of exactly one element in $4\times 6=24ways$.
Thus, the probability of the required event is $24/36=2/3$.