There are four letters and four envelopes and exactly one letter is to be put in exactly one envelop with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements: 1. It is possible that exactly one litter goes into an incorrect envelope. 2. 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?

Correct Option (2) Statement 2 is correct. - To determine the number of ways in which exactly two letters go into their correct envelopes, we first select which two letters are correctly placed. This can be done in $C(4, 2)$ ways. - $C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} =$ 6 ways. - For each of these selections, the remaining two letters must be placed in the remaining two envelopes such that neither is in its correct envelope. This scenario represents a derangement of 2 items. - The number of derangements of 2 items, $D_2$, is 1. For example, if letters $L_3, L_4$ are to be placed in envelopes $E_3, E_4$ incorrectly, the only possible arrangement is $L_3 \to E_4$ and $L_4 \to E_3$. - Therefore, the total number of ways for exactly two letters to be in their correct envelopes is $C(4, 2) \times D_2 = 6 \times 1 = 6$. - The six specific arrangements are: - $L_1 \to E_1, L_2 \to E_2, L_3 \to E_4, L_4 \to E_3$ - $L_1 \to E_1, L_3 \to E_3, L_2 \to E_4, L_4 \to E_2$ - $L_1 \to E_1, L_4 \to E_4, L_2 \to E_3, L_3 \to E_2$ - $L_2 \to E_2, L_3 \to E_3, L_1 \to E_4, L_4 \to E_1$ - $L_2 \to E_2, L_4 \to E_4, L_1 \to E_3, L_3 \to E_1$ - $L_3 \to E_3, L_4 \to E_4, L_1 \to E_2, L_2 \to E_1$ Incorrect Options: Statement 1 is incorrect. - It is not possible for exactly one letter to go into an incorrect envelope. - If exactly one letter is in an incorrect envelope, it implies that the remaining three letters are in their respective correct envelopes. - Consider a scenario where letter $L_1$ is placed in an incorrect envelope, for instance, $E_2$. For only one letter to be incorrect, the remaining three letters ($L_2, L_3, L_4$) must be correctly placed in their respective envelopes ($E_2, E_3, E_4$). - However, this creates a contradiction: $E_2$ is already occupied by $L_1$, making it impossible for $L_2$ to be correctly placed in $E_2$. - Therefore, if one letter is incorrectly placed, at least one other letter must also be incorrectly placed to resolve the displacement or the empty correct envelope. This means a minimum of two letters must be incorrectly placed.

IAS 2023 — CSAT Maths

2023mcqmedium

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

It is possible that exactly one litter 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?

Official previous-year question

Prelims CSAT (Paper II) · Held on 28 May 2023.

Options

A1 only
B2 only
CBoth 1 and 2
DNeither 1 nor 2

Did you get this right?

Solution

Correct Option (2)

Statement 2 is correct.

To determine the number of ways in which exactly two letters go into their correct envelopes, we first select which two letters are correctly placed. This can be done in $C(4, 2)$ ways.
$C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} =$ 6 ways.
For each of these selections, the remaining two letters must be placed in the remaining two envelopes such that neither is in its correct envelope. This scenario represents a derangement of 2 items.
The number of derangements of 2 items, $D_2$ , is 1. For example, if letters $L_3, L_4$ are to be placed in envelopes $E_3, E_4$ incorrectly, the only possible arrangement is $L_3 \to E_4$ and $L_4 \to E_3$ .
Therefore, the total number of ways for exactly two letters to be in their correct envelopes is $C(4, 2) \times D_2 = 6 \times 1 = 6$ .
The six specific arrangements are:
- $L_1 \to E_1, L_2 \to E_2, L_3 \to E_4, L_4 \to E_3$
- $L_1 \to E_1, L_3 \to E_3, L_2 \to E_4, L_4 \to E_2$
- $L_1 \to E_1, L_4 \to E_4, L_2 \to E_3, L_3 \to E_2$
- $L_2 \to E_2, L_3 \to E_3, L_1 \to E_4, L_4 \to E_1$
- $L_2 \to E_2, L_4 \to E_4, L_1 \to E_3, L_3 \to E_1$
- $L_3 \to E_3, L_4 \to E_4, L_1 \to E_2, L_2 \to E_1$

Incorrect Options:

Statement 1 is incorrect.

It is not possible for exactly one letter to go into an incorrect envelope.
If exactly one letter is in an incorrect envelope, it implies that the remaining three letters are in their respective correct envelopes.
Consider a scenario where letter $L_1$ is placed in an incorrect envelope, for instance, $E_2$ . For only one letter to be incorrect, the remaining three letters ( $L_2, L_3, L_4$ ) must be correctly placed in their respective envelopes ( $E_2, E_3, E_4$ ).
However, this creates a contradiction: $E_2$ is already occupied by $L_1$ , making it impossible for $L_2$ to be correctly placed in $E_2$ .
Therefore, if one letter is incorrectly placed, at least one other letter must also be incorrectly placed to resolve the displacement or the empty correct envelope. This means a minimum of two letters must be incorrectly placed.

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Solution

Correct Option (2)

Statement 2 is correct.

To determine the number of ways in which exactly two letters go into their correct envelopes, we first select which two letters are correctly placed. This can be done in $C(4, 2)$ ways.
$C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} =$ 6 ways.
For each of these selections, the remaining two letters must be placed in the remaining two envelopes such that neither is in its correct envelope. This scenario represents a derangement of 2 items.
The number of derangements of 2 items, $D_2$ , is 1. For example, if letters $L_3, L_4$ are to be placed in envelopes $E_3, E_4$ incorrectly, the only possible arrangement is $L_3 \to E_4$ and $L_4 \to E_3$ .
Therefore, the total number of ways for exactly two letters to be in their correct envelopes is $C(4, 2) \times D_2 = 6 \times 1 = 6$ .
The six specific arrangements are:
- $L_1 \to E_1, L_2 \to E_2, L_3 \to E_4, L_4 \to E_3$
- $L_1 \to E_1, L_3 \to E_3, L_2 \to E_4, L_4 \to E_2$
- $L_1 \to E_1, L_4 \to E_4, L_2 \to E_3, L_3 \to E_2$
- $L_2 \to E_2, L_3 \to E_3, L_1 \to E_4, L_4 \to E_1$
- $L_2 \to E_2, L_4 \to E_4, L_1 \to E_3, L_3 \to E_1$
- $L_3 \to E_3, L_4 \to E_4, L_1 \to E_2, L_2 \to E_1$

Incorrect Options:

Statement 1 is incorrect.

It is not possible for exactly one letter to go into an incorrect envelope.
If exactly one letter is in an incorrect envelope, it implies that the remaining three letters are in their respective correct envelopes.
Consider a scenario where letter $L_1$ is placed in an incorrect envelope, for instance, $E_2$ . For only one letter to be incorrect, the remaining three letters ( $L_2, L_3, L_4$ ) must be correctly placed in their respective envelopes ( $E_2, E_3, E_4$ ).
However, this creates a contradiction: $E_2$ is already occupied by $L_1$ , making it impossible for $L_2$ to be correctly placed in $E_2$ .
Therefore, if one letter is incorrectly placed, at least one other letter must also be incorrectly placed to resolve the displacement or the empty correct envelope. This means a minimum of two letters must be incorrectly placed.

IAS 2023 — CSAT Maths

Options

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Solution

Correct Option (2)

Incorrect Options:

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IAS 2023 — CSAT Maths

Options

Did you get this right?

Solution

Correct Option (2)

Incorrect Options:

Your note

IAS Maths in other years

More IAS Maths questions

Other CSAT topics for IAS

Comprehension

Reasoning

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