IPMAT Indore 2022 · Quantitative Ability · Question 28

Step 1 — count the relations. A relation from A to B is any subset of A\times B. Since

\left| A\times B \right|=3\times 2=6

the number of relations is

{2}^{6}=64

Step 2 — count the functions. A function assigns to each element of A exactly one element of B. With 3 elements in A and 2 choices each:

{2}^{3}=8

Step 3 — divide.

P=\frac{8}{64}=\frac{1}{8}

Answer: Option 1 — \frac{1}{8}

Shortcut. Two formulas, both worth knowing cold:

\text{relations}A\to B={2}^{\left| A \right|\left| B \right|}, \text{functions}A\to B={\left| B \right|}^{\left| A \right|}

So the probability is

\frac{{\left| B \right|}^{\left| A \right|}}{{2}^{\left| A \right|\left| B \right|}}=\frac{{2}^{3}}{{2}^{6}}=\frac{1}{{2}^{3}}=\frac{1}{8}

Option (d) {3}^{2}/{2}^{6} is the trap: it swaps the roles of \left| A \right| and \left| B \right| in the numerator, giving {3}^{2} instead of {2}^{3}. The base is the size of the codomain — each element of A independently picks one of \left| B \right| targets.