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Consider all distinguishable arrangements of the multiset of letters…
Question
Consider all distinguishable arrangements of the multiset of letters {M, I, I, I, I, S, S, S, S, P, P} (one M, four I's, four S's, two P's). How many of these arrangements have no two S's adjacent?
✓ Verified answer: 7350checked by our engine — not a guess
Step-by-step solution
First arrange the seven non-S letters: one M, four I's, two P's. The number of distinguishable arrangements is 7!/(4!·2!) = 5040/48 = 105.
These 7 letters create 8 gaps (including the two ends) into which the four identical S's can be placed, at most one S per gap to guarantee no two S's are adjacent.
Choose 4 of the 8 gaps: C(8,4) = 70.
Total = 105 * 70 = 7350.
Final answer7350
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