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 ?

The correct option is 6.

To find the number of different four-digit numbers that can be formed using the digits 1, 2, 3, and 4 without repetition, let us analyze the restrictions step-by-step.

Without any restrictions, the total number of permutations of 4 distinct digits is given by:
$4!=4\times 3\times 2\times 1=24$
We can find the valid numbers by examining the possible starting digits based on the rules.

Rule 1: 2 and 3 cannot be next to each other (i.e., we cannot have "23" or "32" in the number).
Rule 2: 1 cannot be immediately followed by 3 (i.e., we cannot have "13" in the number).
Rule 3: 4 cannot be at the last place.
Rule 4: 1 cannot be at the first place.

Since 1 cannot be at the first place, the first digit must be 2, 3, or 4. Let us analyze each case:

Case 1: First digit is 2
The remaining digits to place are 1, 3, and 4. The possible arrangements and their validity are:
- 2134: Invalid (contains "13" and has 4 at the last place)
- 2143: Valid (no adjacent 2-3/3-2, no "13", last digit is 3, first is 2)
- 2314: Invalid (contains "23")
- 2341: Invalid (contains "23")
- 2413: Invalid (contains "13")
- 2431: Valid (no adjacent 2-3/3-2, no "13", last digit is 1, first is 2)
There are 2 valid numbers in this case: 2143 and 2431.

Case 2: First digit is 3
The remaining digits to place are 1, 2, and 4. The possible arrangements and their validity are:
- 3124: Invalid (4 is at the last place)
- 3142: Valid (no adjacent 2-3/3-2, no "13", last digit is 2, first is 3)
- 3214: Invalid (contains "32")
- 3241: Invalid (contains "32")
- 3412: Valid (no adjacent 2-3/3-2, no "13", last digit is 2, first is 3)
- 3421: Valid (no adjacent 2-3/3-2, no "13", last digit is 1, first is 3)
There are 3 valid numbers in this case: 3142, 3412, and 3421.

Case 3: First digit is 4
The remaining digits to place are 1, 2, and 3. Since 4 is at the first place, it cannot be at the last place. The possible arrangements and their validity are:
- 4123: Invalid (contains "23")
- 4132: Invalid (contains "13")
- 4213: Invalid (contains "13")
- 4231: Invalid (contains "23")
- 4312: Valid (no adjacent 2-3/3-2, no "13", last digit is 2, first is 4)
- 4321: Invalid (contains "32")
There is 1 valid number in this case: 4312.

Adding the valid combinations from all cases:
$2+3+1=6$
Thus, exactly 6 different numbers can be formed under the given constraints.