Permutation and Combination

Treat A and B as a single unit. There are 3 units (AB, C, D) to arrange in 3! = 6 ways. Within the unit, A and B can be arranged in 2! = 2 ways. Total = 6 * 2 = 12.

Arrange 6 '+' signs: _ + _ + _ + _ + _ + _ _. There are 7 gaps. We need to place 4 '-' signs in these 7 gaps such that no two are together. This is 7C4 = 35 ways.

To select 6 cards with at least one from each suit, the distribution of suits can be (3,1,1,1) or (2,2,1,1). For (3,1,1,1), choose 1 suit for 3 cards (4C1), choose 3 cards from it (13C3), choose 1 card from each of the other 3 suits (13C1^3). For (2,2,1,1), choose 2 suits for 2 cards (4C2), choose 2 cards from each (13C2^2), choose 1 card from each of the other 2 suits (13C1^2).

To be even, the last digit must be 4 or 8. If the last digit is 4, there are 4 choices for the first, 3 for the second, 2 for the third, and 1 for the fourth (4 * 3 * 2 * 1 = 24). If the last digit is 8, there are also 24 ways. Total = 24 + 24 = 48.

RECOGNISE has 9 letters: 4 vowels (E, O, I, E) and 5 consonants (R, C, G, N, S). Vowels must be at even positions (2, 4, 6, 8). There are 4 even positions for 4 vowels. Number of arrangements = (4! / 2!) * 5! = 12 * 120 = 1440.

Total arrangements of 6 players = 6! = 720. Arrangements where Ajeet and Mukherjee are together: treat them as one unit, so 5! * 2! = 120 * 2 = 240. Arrangements where they are not together = 720 - 240 = 480.

JUBILANCE has 9 letters: J, U, B, I, L, A, N, C, E. Consonants: J, B, L, N, C (5). Vowels: U, I, A, E (4). First letter can be any of 5, last any of 4. Remaining 7 letters can be arranged in 7! ways. 5 * 4 * 5040 = 100,800.

The word THURSDAY has 8 distinct letters. The number of permutations of n distinct items is n!, which is 8!.

Total arrangements of BHARAT (6 letters, 2 A's) = 6! / 2! = 360. Arrangements where B and H are together (treat BH as one unit) = 5! / 2! = 60. Since B and H can be arranged in 2 ways (BH or HB), total together = 60 * 2 = 120. Never together = 360 - 120 = 240.