2023 CSAT — Q6

Quant Statement validity 2.5 marks Medium

There are four letters and four envelopes and exactly one letter is to be put in exactly one envelope with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements:

It is possible that exactly one letter goes into an incorrect envelope.
There are only six ways in which only two letters can go into the correct envelopes.

Which of the statements given above is/are correct?

A 1 only
B 2 only Answer
C Both 1 and 2
D Neither 1 nor 2

Worked rationale

The whole item turns on the fixed-point structure of a permutation of $4$ letters into $4$ envelopes.

Statement 1 — “exactly one letter goes into an incorrect envelope” means exactly one is wrong and the other three are correct. But if three letters sit in their correct envelopes, the fourth has only its own envelope left, so it is forced correct too. You can never have exactly one wrong (a permutation cannot have exactly one non-fixed point — there is no derangement of a single element). Statement 1 is false.

Statement 2 — “only two letters go into the correct envelopes” means exactly two are correct and the other two are both wrong. Choose which two are correct: $\binom{4}{2} = 6$ ways. The remaining two letters must both be misplaced — they must swap, which is the only derangement of two elements ( $D_2 = 1$ ). So the count is $6 \times 1 = 6$ . Statement 2 is true.

Answer: (b) 2 only.

Why the other options miss

A

misses the forcing argument: accepts “exactly one wrong” as possible, missing that the other three being correct forces the fourth correct (no permutation has a lone non-fixed point).
C

gets Statement 2 right but still admits Statement 1, failing the “you cannot derange exactly one element” check.
D

miscounts the exactly-two-correct case (e.g. forgets the remaining two must swap, or double-counts), rejecting the true Statement 2.

Specialist insight

The killer idea is the impossibility of exactly one non-fixed point: a derangement of one element does not exist, so “exactly $k$ wrong” is impossible for $k = 1$ . Once that lands, Statement 2 is a one-line “choose the two correct ( $\binom{4}{2}$ ), force the other two to swap ( $D_2 = 1$ )” count. The general exactly- $k$ -correct formula is $\binom{n}{k} \cdot D_{n-k}$ ; here $\binom{4}{2}D_2 = 6 \cdot 1 = 6$ .

The trap, in one line

You can never have exactly one letter wrong (three correct forces the fourth), so Statement 1 is false; exactly two correct $=\binom{4}{2}\cdot D_2 = 6$ .

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