Permutation and Combination

LEADER has six letters, with E repeated twice. The total number of distinct arrangements is 6!/2! = 360, and arrangements with adjacent Es number 5! = 120, so the required count is 360 - 120 = 240.

If one person is always included, we need to choose 3 more from the remaining 9. Ways = 9C3 = (9 * 8 * 7) / (3 * 2 * 1) = 3 * 4 * 7 = 84.

Arranging 12 flowers in a garland: (n-1)! / 2. Treating 4 flowers as one unit, we have 9 units (8 + 1). Arrangements = (9-1)! / 2 = 8! / 2 = 40320 / 2 = 20160. Within the unit, the 4 flowers can be arranged in 4! = 24 ways. Total = 20160 * 24 = 483840.

The word SUNDAY has 6 distinct letters. The number of arrangements is the factorial of the number of letters, which is 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.

To form a word with 2 vowels and 3 consonants, first select them: 5C2 ways for vowels and 6C3 ways for consonants. Then, arrange the 5 selected letters in 5! ways. The total is 5C2 * 6C3 * 5!.

Total questions = 12. Must choose 8. Questions 1 and 3 are fixed, so we must choose 6 more from the remaining 10. Number of ways = 10C6 = 10C4 = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 210.

The number of permutations of 'Mississippi' is 11! / (4! * 4! * 2!), where 4 is the count of 'i', 4 is the count of 's', and 2 is the count of 'p'. The formula provided is correct.

The word 'INFINITE' has 8 letters: I(3), N(2), F(1), T(1), E(1). Combinations of 4 letters can be: 3 of one kind + 1 of another (2 cases: I+others), 2 of one kind + 2 of another (1 case: I,N), 2 of one kind + 2 different (3 cases: I,N), or 4 different (4 cases). Summing these yields 22.

Number of ways to choose 2 books from each group: C(5,2) * C(4,2) * C(3,2) = 10 * 6 * 3 = 180. These 6 chosen books can be arranged in 6! ways. Total arrangements = 180 * 6!.

This is a combinatorics problem. We have 20 positions. We must choose 10 positions for the x-series and 10 for the y-series. Once positions are chosen, the order is fixed (ascending). The number of ways is 20C10.

ARTICLE has 7 letters: 3 vowels (A, I, E) and 4 consonants (R, T, C, L). Odd positions are 1, 3, 5, 7 (4 spots). Vowels must be in odd positions: 4P3 = 4 * 3 * 2 = 24 ways. Remaining 4 letters in 4 spots: 4! = 24 ways. Total = 24 * 24 = 576.

Total possible 5-letter words from 10 letters = 10^5 = 100,000. Words with no letters replaced (all distinct) = 10*9*8*7*6 = 30,240. Words with at least one letter repeated (replaced) = 100,000 - 30,240 = 69,760.