This can be done in $\binom{22}{5} = 26,\!296$ ways.
Now, let's find the number of ways to form a committee with no women or only one woman. There are $\binom{10}{5} = 252$ ways to form a committee with no women, and $\binom{10}{4} \cdot \binom{12}{1} = 7,\!290$
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