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You should use permutation and combination as follow:-

You have $6$ length bit string in which you want more no of zeros than ones.

So all possible combination are as follows-

$4$ zeros and $2$ one

$5$ zeros and $1$ one

$6$ zeros and $0$ one

C(6,4) + C(6,5) + C(6,6).

This would be-

$15 + 6 + 1$

Ans = $22$.

So, there are $22$ such strings.
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