Permutation and Combination

The number of ways to arrange n distinct objects is n!. For 6 books, it is 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.

The word MANGO has 5 letters: M, A, N, G, O. There are 5! = 120 total arrangements. The vowels are A and O. If the word starts with a vowel, the first position can be filled in 2 ways (A or O), and the remaining 4 positions in 4! = 24 ways. Total favorable arrangements = 2 * 24 = 48. Probability = 48 / 120 = 2/5.

Total arrangements of MATRIX (6 letters) is 6! = 720. Vowels are A and I. Treating (AI) as one unit, we have 5 units (M, T, R, X, AI), which can be arranged in 5! * 2! = 240 ways. Subtracting this from 720 gives 480.

The number of ways to arrange 3 books out of 5 is given by the permutation formula P(n, r) = n! / (n-r)!. Here, P(5, 3) = 5 * 4 * 3 = 60.

The word 'BALLOONS' has 8 letters: B, A, L, L, O, O, N, S. L and O repeat twice. Total arrangements = 8! / (2! * 2!) = 40320 / 4 = 10080.

Since 4 members of the 7-person task force have already been selected from the 12 candidates, the supervisor needs to choose 3 more members from the remaining 8 candidates. The number of ways to do this is given by the combination formula 8C3 = (8 * 7 * 6) / (3 * 2 * 1) = 56.

OMEGA has 5 letters: 3 consonants (M, G, N? No, O, M, E, G, A). Vowels are O, E, A. Consonants are M, G. Odd places are 1, 3, 5. There are 3 odd places and 3 vowels. The number of ways to arrange the vowels in odd places is 3! = 6. The remaining 2 consonants can be arranged in the 2 even places in 2! = 2 ways. Total = 6 * 2 = 12.

Choose 4 consonants from 7: 7C4 = 35. Choose 2 vowels from 5: 5C2 = 10. Total combinations = 35 * 10 = 350. Arrange 6 letters: 6! = 720. Total = 350 * 720 = 252,000.

Treat the Hindi, Math, and Science books as one single unit. There are now 8 units (7 individual books + 1 group of 3). These 8 units can be arranged in 8! ways. Within the group, the 3 books can be arranged in 3! ways. Total arrangements = 8! * 3!.

MATHEMATICS has 11 letters: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1). Vowels are A, E, A, I (4 vowels). Treating them as one block, we have 8 units (7 consonants + 1 vowel block). Arrangements = (8! / (2! * 2!)) * (4! / 2!) = (40320 / 4) * (24 / 2) = 10080 * 12 = 120,960.

Each of the 10 boys can either be in group 1 or group 2. This gives 2^10 = 1024 total assignments. Excluding the cases where one group is empty (2 cases), we have 1022. Since the order of groups does not matter, we divide by 2, resulting in 511.

SCHOOL has 6 letters: S, C, H, O, O, L. Consonants: S, C, H, L (4). Vowels: O, O (2). Arrange consonants: 4! = 24. Spaces for vowels: _ S _ C _ H _ L _. 5 spaces for 2 vowels. Arrangements = 24 * 5P2 / 2! = 24 * 10 = 240.

Total permutations of DAUGHTER (8 letters, all distinct) = 8! = 40320. Vowels are A, U, E (3). Treat (AUE) as one unit. Permutations with vowels together = 6! * 3! = 720 * 6 = 4320. Vowels never together = 40320 - 4320 = 36000.