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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?
[2016]
  • a)
    6
  • b)
    8
  • c)
    9
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?
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Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; an...
Problem: How many different four-digit numbers can be formed using the digits 1, 2, 3 and 4, such that none of these digits are repeated in any manner and the given conditions are satisfied?

Solution:

We consider each condition one by one:

Condition 1: 2 and 3 are not to immediately follow each other
We can place 2 and 3 in two ways:
- 2 _ 3
- 3 _ 2

In the first case, we can place 1 and 4 in any of the remaining two places. So, there are 2 ways to place 2 and 3 and 2 ways to place 1 and 4. Therefore, there are 2 × 2 = 4 ways to place 1, 2, 3, and 4.

In the second case, we can place 1 in the first place and 4 in the last place. So, the remaining two places can be filled in 2! = 2 ways using 2 and 3. Therefore, there are 2 ways to place 2 and 3 and 2 ways to place 1 and 4. Therefore, there are 2 × 2 = 4 ways to place 1, 2, 3, and 4.

Total number of ways = 4 + 4 = 8

Condition 2: 1 is not to be immediately followed by 3
We can place 1 in the first place and 4 in the last place. So, the remaining two places can be filled in 2! = 2 ways using 2 and 3. Therefore, there are 2 ways to place 1 and 4 and 2 ways to place 2 and 3. Therefore, there are 2 × 2 = 4 ways to place 1, 2, 3, and 4.

Total number of ways = 4

Condition 3: 4 is not to appear at the last place
We can place 4 in any of the three places. So, there are 3 ways to place 4. For each of these ways, we can place 1 in the first place in 3 ways (excluding the case where 3 is in the second place), and then the remaining two places can be filled in 2! = 2 ways using 2 and 3. Therefore, there are 3 × 3 × 2 = 18 ways to place 1, 2, 3, and 4.

Condition 4: 1 is not to appear at the first place
We can place 1 in any of the three places (excluding the first place). So, there are 3 ways to place 1. For each of these ways, we can place 4 in the last place in 1 way. Then, the remaining two places can be filled in 2! = 2 ways using 2 and 3. Therefore, there are 3 × 1 × 2 = 6 ways to place 1, 2, 3, and 4.

Total number of ways = 6

Therefore, the total number of different numbers that can be formed = 6 (Option A).
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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 other2. 1 is not to be immediately followed by 33. 4 is not to appear at the last place4. 1 is not to appear at the first placeHow many different numbers can be formed?[2016]a)6b)8c)9d)None of the aboveCorrect answer is option 'A'. Can you explain this answer?
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