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how many bit strings of length 10 contain either five consecutive 0's or five consecutive 1's ?

Sagot :

The first five 0s can start at position 1, 2, 3, 4, or 6 if you follow the first five 0s summation rule.

Start with the first slot then add anything else after that. Starting at position 2, the first bit must be a 1 (2'5 = 32).

the remaining bits can be anything; 2'4 = 16 start at third spot, the second bit must be a 1 owing to the reason given above, the first bit can be whatever, and the last 3 bits can be anything;

16 starting at spots 4,5 and 6, the same as starting at spots 2 or 3, for a total of 16 each. 32 + 16 + 16 + 16 + 16 + = 112.

The same pattern is guaranteed by the five straight 1s, and there are several possibilities, such as 112.

A double count would apply to two cases (thus that we need to exclude)

0000011111 and 1111100000

Total = 112 +112 -2 =222

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