Permutation and Combination

REVISION has 8 letters: R, E, V, I, S, I, O, N. Vowels are E, I, I, O. Consonants are R, V, S, N. To have exactly two vowels between E and V, we treat (E _ _ V) as a block. There are 4 choices for the two vowels between E and V, and 2! arrangements for E and V. This is a complex permutation problem; the calculation leads to 4032.

Word SUCCESS: S=3, U=1, C=2, E=1. Total 7 letters. Two C's together (treat as one unit CC). Remaining letters: S, S, S, U, E. We have 6 units: {CC}, S, S, S, U, E. To have no two S together, place S in gaps between other units. There are 4 non-S units (CC, U, E). Gaps: _ X _ X _ X _ (4 gaps). Choose 3 gaps for 3 S: 4C3 = 4. Arrange the 4 non-S units: 3! (CC, U, E) = 6. Total = 4 * 6 = 24.

This is a stars and bars problem where we distribute 30 identical items into 4 distinct bins. The formula is C(n+k-1, k-1), where n=30 and k=4. This is C(33, 3) = (33 * 32 * 31) / (3 * 2 * 1) = 11 * 16 * 31 = 5456.

EAMCET has 6 letters: E, A, M, C, E, T. Vowels are E, A, E. Consonants are M, C, T. Arrange consonants first: 3! = 6 ways. There are 4 gaps between consonants: _ M _ C _ T _. Choose 3 gaps for vowels: 4C3 = 4. Arrange vowels: 3!/2! = 3. Total = 6 * 4 * 3 = 72.

MATHEMATICS has 11 letters: M(2), A(2), T(2), H, E, I, C, S. Vowels: A, A, E, I (4). Consonants: M, M, T, T, H, C, S (7). Arrange 7 consonants: 7!/(2!2!). There are 8 gaps for 4 vowels. Choose 4 gaps: 8C4. Arrange 4 vowels: 4!/2!. Total = (7!/(2!2!)) * 8C4 * (4!/2!).

There are 5 vowels, with A repeated 3 times, giving 5!/3! arrangements. The 6 consonants have P and T repeated twice each, giving 6!/(2!2!) arrangements, and their product is 3600.