Consider all permutations of the word MAHAKUMBH. Let N be the number of such permutations in which no two identical letters appear adjacent to each other. Find the total number of divisors of N.

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Accepted Answer

48. Letters: M(2), A(2), H(2), K, U, B. Total = 9!/(2!*2!*2!) = 362880/8 = 45360. Using inclusion-exclusion: |X| = 8!/(2!*2!) = 10080 (treat MM as one unit, remaining letters A,A,H,H,K,U,B). Similarly |Y| = |Z| = 10080. |X inter Y| = 7!/2! = 2520 (MM and AA as units). |X inter Z| = |Y inter Z| = 2520. |X inter Y inter Z| = 6! = 720. |X union Y union Z| = 3(10080) - 3(2520) + 720 = 30240 - 7560 + 720 = 23400. N = 45360 - 23400 = 21960. Factorization: 21960 = 2³ * 3² * 5 * 61. Number of divisors = (3+1)(2+1)(1+1)(1+1) = 4*3*2*2 = 48.

Consider all permutations of the word MAHAKUMBH. Let N be the number of such permutations in which no two identical letters appear adjacent to each other. Find the total number of divisors of N.

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