Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these four digits are repeated in any manner. Further, 1. 2 and 3 are not to immediately follow each other. 2. 1 is not to be immediately followed by 3. 3. 4 is not to appear at the last place. 4. 1 is not to appear at the first place. How many different numbers can be formed?

QUESTION

CSAT

Hard

Maths

Prelims 2016

Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these four digits are repeated in any manner. Further,

2 and 3 are not to immediately follow each other.
1 is not to be immediately followed by 3.
4 is not to appear at the last place.
1 is not to appear at the first place.

How many different numbers can be formed?

Select an option to attempt

Explanation

We need to form four-digit numbers using the digits 1, 2, 3, and 4, ensuring:

2 and 3 do not appear consecutively 1 is not immediately followed by 3 4 does not appear in the last place 1 does not appear in the first place

Step-by-Step Calculation: Step 1: Total Arrangements Since all four digits are distinct, the total number of ways to arrange them is: $4! = 24$

Step 2: Eliminating Invalid Cases We count and remove cases violating the given conditions. 1 not in the first position: If 1 is fixed in the first position, the remaining three digits can be arranged in: $3! = 6$ (invalid cases)

Valid cases left: $24 - 6 = 18$

4 not in the last position: If 4 is fixed in the last position, remaining three digits can be arranged in: $3! = 6$ (invalid cases) Valid cases left: $18 - 6 = 12$ 2 and 3 not together + 1 not followed by 3: