Question 257542
(This involves permutations.)Suppose that the 10 red balls are labeled as R1, R2, ..., R10 and they are randomly arranged in a single line. What is the probability that R1, R2 and R3 are in adjacent positions?
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Since the 3 must place together
there are only 8! arranements.
But the 3 can be arranged among themselves in 3! = 6 ways.
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Total arrangements 8!*6 = 241920 arrangements.
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Cheers,
Stan H.