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.