CSAT Solved Papers/ 2022/Q24

2022 CSAT — Q24

Quant Counting & combinatorics 2.5 marks Medium

The letters $A, B, C, D$ and $E$ are arranged in such a way that there are exactly two letters between $A$ and $E$ . How many such arrangements are possible?

A 12
B 18
C 24 Answer
D 36

Worked rationale

Five positions $1$ – $5$ . “Exactly two letters between $A$ and $E$ ” means their positions differ by $3$ . The position-pairs with a gap of $3$ are $\{1,4\}$ and $\{2,5\}$ — two unordered slot-pairs.

For each slot-pair, $A$ and $E$ can be in either order: $2$ ways. So placements of $(A,E)$ :

2 \text{ slot-pairs} \times 2 \text{ orders} = 4.

The remaining three letters $B, C, D$ fill the other three positions in $3! = 6$ ways. Total:

4 \times 6 = 24.

Answer: (c) 24.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2022 CSAT Q24 — Counting & combinatorics — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

dropped the order factor: counts only one order of $A,E$ (forgets $A$ – $E$ vs $E$ – $A$ ), halving the count.
B

wrong slot-pair count: mixes a bad slot-pair tally ( $3$ ) with $3!$ , or mis-multiplies.
D

too many slot-pairs allowed: over-counts the slot-pairs (e.g. lets gap-of- $2$ pairs in too), inflating to $6 \times 6$ .

Specialist insight

“Exactly two letters between” $\Rightarrow$ a positional gap of $3$ — the cleanest reading, and the only gap that yields two slot-pairs in five seats: $\{1,4\}$ and $\{2,5\}$ . Count the constrained pair first ( $\times 2$ for order), then the free letters ( $3!$ ). The trap is dropping the order factor (giving $12$ ) or miscounting the slot-pairs.

The trap, in one line

Gap- $3$ slot-pairs $\{1,4\},\{2,5\}$ ; $2\times 2$ placements of $(A,E)$ times $3!$ $= 24 \Rightarrow$ (c).

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