Permutation and Combination

The word TRICK has 5 distinct letters. If C is fixed in the middle, we have 4 positions left for the remaining 4 letters (T, R, I, K). The number of arrangements is 4! = 4 * 3 * 2 * 1 = 24.

Since the mangoes are identical, we can use the stars and bars method to distribute 30 identical items among 4 distinct boys. This is equivalent to finding the number of non-negative integer solutions to x1 + x2 + x3 + x4 = 30, which is given by the combination formula C(n + r - 1, r - 1) = C(30 + 4 - 1, 4 - 1) = C(33, 3).

Each of the 4 items has 3 choices of boxes. Since the items are distinct and boxes are distinct, the total ways = 3 * 3 * 3 * 3 = 3^4.

First arrange the 7 non-S letters, giving 7!/(4! x 2!) arrangements. These letters create 8 gaps, and choosing 4 gaps for the S letters ensures that no two S letters are adjacent, giving 7!/(4! x 2!) x C(8,4).

Total arrangements of BHARAT (6 letters, 2 A's) = 6! / 2! = 360. Arrangements where B and H are together: Treat (BH) as one unit. Arrangements of (BH), A, R, A, T = 5! / 2! = 60. Since BH can be HB, multiply by 2 = 120. Total - Together = 360 - 120 = 240.

The word TRIANGLE has 8 distinct letters, 3 vowels (A, I, E) and 5 consonants (T, R, N, G, L). To keep vowels apart, arrange the 5 consonants first (5! = 120 ways) and place the 3 vowels in the 6 available gaps (6P3 = 6 * 5 * 4 = 120 ways). Total = 120 * 120 = 14400.

Total permutations of BHARAT (6 letters, A repeats twice) is 6! / 2! = 360. Permutations where B and H are together (treat BH as one unit) is 5! / 2! = 60. Since BH can be HB, multiply by 2 to get 120. Subtracting 120 from 360 gives 240.

The word HOTEL has 5 letters: H, O, T, E, L. Vowels are O and E. Treat (OE) as one unit. We have 4 units: H, T, L, (OE). These can be arranged in 4! = 24 ways. The vowels (OE) can be arranged in 2! = 2 ways. Total = 24 * 2 = 48.

The student must choose 4 questions out of 5. Questions 1 and 2 are compulsory, so the student must choose 2 more questions from the remaining 3 (5 - 2 = 3). The number of ways is 3C2 = 3.

Total possible 4-letter words using 10 letters is 10^4 = 10000. Words with no repeated letters is 10 * 9 * 8 * 7 = 5040. The number of words with at least one repetition is 10000 - 5040 = 4960.