ISRO CS 2025 Q.61: solution and answer

ISRO 2025 · CSQ.61 · 1 mark · MCQ

If $A$ is a set with $m$ elements and $n$ is a positive integer, the number of different elements in $A^{n}$ is :

(A)
$m \times n$
(B)
$m \times m$
(C)
$n^{m}$
(D)
$m^{n}$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The number of elements in $A^{n}$ is $m^{n}$ , option (D).

Official answer: DSource: ISRO's final answer key of 15 May 2026

$A^{n}$ is the $n$ -fold Cartesian product $A \times A \times \dots \times A$ , so each element is an ordered $n$ -tuple $(a_{1}, a_{2}, \dots, a_{n})$ with every $a_{i} \in A$ .
By the multiplication rule of counting, each of the $n$ positions can be filled independently in $∣ A ∣ = m$ ways.
Hence $∣ A^{n} ∣ = n factors m \times m \times \dots \times m = m^{n}$ .
Check with $n = 2$ : $∣ A \times A ∣ = m \cdot m = m^{2}$ , which agrees with $m^{n}$ and not with $m \times n = 2 m$ (the two are equal only when $m = 2$ ).
Option (a) $m \times n$ adds the sizes instead of multiplying per position. Option (b) $m \times m$ is only the case $n = 2$ . Option (c) $n^{m}$ swaps the base and the exponent: it counts functions from an $m$ -set to an $n$ -set.

KEY IDEA

$A^{n}$ is the set of ordered $n$ -tuples over $A$ . Each of the $n$ positions has $m$ independent choices, so the count is $m \times m \times \dots \times m = m^{n}$ .

WATCH OUT FOR

Swapping base and exponent to get $n^{m}$ , or treating $A^{n}$ as $m \times n$ (adding sizes instead of multiplying choices per position).

From the ISRO CS 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 5 ISRO Counting & axioms questions, with notes

ISRO CS 2025 Q.61: solution and answer

Answer and solution

More ISRO Counting & axioms questions