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Basically, this is the same as finding the number of distinct ways of arranging seven 1s and 3 0s. That is (10!/7!3!) = (10*9*8)/(3*2*1) = 120.

There are 120 bit strings of length 10 with exactly three 0s.

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Q: How many bit strings of length 10 have exactly three 0s?
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