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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56 The number of triples of 1s on 8 bits
There are 210.
75 i think
Every number from 100 to 999 inclusive !
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