Will the three A's clump together?

Question

You have six lettered tiles, three marked A, two marked B, and one marked C, and you lay them out in a random left-to-right row. Tiles with the same letter are indistinguishable, so two rows count as "the same" only if they read identically.

What is the probability that the three A's end up all together as one consecutive block?

Accepted Answer

First count all distinct rows. With three identical A's and two identical B's among six tiles, the number of distinguishable arrangements is \frac{6!}{3!\,2!\,1!} = \frac{720}{12} = 60. Now count the arrangements where the A's form a single block. Glue the three A's into one super-tile "AAA". We are then arranging four items: the AAA block, B, B, and C. The number of distinguishable orders of those is \frac{4!}{2!} = 12, dividing by 2! for the two identical B's. So the probability is \frac{12}{60} = \frac{1}{5} = 0.2. Takeaway. To force a group of items to stay adjacent, treat them as one glu…

Will the three A's clump together?

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