Let X be the set of all distinct words formed using all letters of SHREYANSH and Y be the set of all distinct words formed using all letters of SANIDHYA. A set is chosen at random (each with probability 1/2) and then a word is selected at random from that set. If the probability that the s

Question

Let X be the set of all distinct words formed using all letters of SHREYANSH and Y be the set of all distinct words formed using all letters of SANIDHYA. A set is chosen at random (each with probability 1/2) and then a word is selected at random from that set. If the probability that the selected word contains at least one pair of identical adjacent letters is p/q (where p and q are coprime positive integers), find the value of (q - 3p).

Accepted Answer

3. SHREYANSH: S(2), H(2), R(1), E(1), Y(1), A(1), N(1) — 9 letters total. Total arrangements = 9!/(2!*2!) = 90720. SANIDHYA: S(1), A(2), N(1), I(1), D(1), H(1), Y(1) — 8 letters total. Total = 8!/2! = 20160. For SHREYANSH (repeated: 2S, 2H), using complementary: arrangements with no two S adjacent AND no two H adjacent. For SANIDHYA (repeated: 2A): arrangements with no two A adjacent = 8!/2! - 7! = 20160 - 5040 = 15120. P(adjacent pair | SANIDHYA) = 1 - 15120/20160 = 1 - 3/4 = 1/4. For SHREYANSH: arrangements with SS together OR HH together = (arrange 8 distinct-like with SS as one unit: 8!/2! for H's) + same for HH - both. This requires careful inclusion-exclusion. The combined probability

Solution

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