Question

Consider all functions that can be defined from the set $A=\{1,2,3\}$ to the set $B=\{1,2,3,4,5\}$. A function is selected at random from these functions. The probability that selected function satisfies $f\left(i\right)\le f\left(j\right)$, for $i<j$ is equal to

• A. $\frac{6}{25}$
• B. $\frac{7}{25}$
• C. $\frac{2}{25}$
• D. $\frac{12}{25}$

Answer 1

Answer: (B)

$\frac{7}{25}$

Set $A$ has $3$ elements and set $B$ has $5$ elements.

While mapping from $A$ to $B$, therefore, there can be $5^3=125$ functions.
In order to satisfy the condition:

$f\left(i\right)\le f\left(j\right)$ for $i<j$.

1. If $f\left(1\right)=5$, then $f\left(2\right)=f\left(3\right)=5$ i.e. only $1$ function exists
2. If $f\left(1\right)=4$, then $f\left(2\right)=4,f\left(3\right)=4$ or $f\left(2\right)=4,f\left(3\right)=5$ or $f\left(2\right)=5,f\left(3\right)=5$ i.e. only $3$ functions exist
3. If $f\left(1\right)=3$, then $f\left(2\right)=3,f\left(3\right)=3$ or $f\left(2\right)=3,f\left(3\right)=4$ or $f\left(2\right)=3,f\left(3\right)=5$ or $f\left(2\right)=4,f\left(3\right)=4$ or $f\left(2\right)=4,f\left(3\right)=5$ or $f\left(2\right)=5,f\left(3\right)=5$ i.e. only $6$ functions exist
Similarly, for
4. If $f\left(1\right)=2$, there would be $10$ functions
and
5. If $f\left(1\right)=1$, there would be $15$ functions

Thus, there are a total of $1+3+6+10+15=35$ possible functions satisfying the given condition.

Hence, the required probability = $\frac{35}{125}=\frac{7}{25}$.