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In how many ways can a committee of 5 made out 6 men and 4 women containing atleast one woman?
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The correct answer is:
246
A committee of 5 out of $6+4=10$ can be made in ${ }^{10} \mathrm{C}_{5}=252$ ways. If no woman is to be included, thennumber of ways $={ }^{5} \mathrm{C}_{5}=6$
$\therefore$ the required number $=252-6=246$
$\therefore$ the required number $=252-6=246$
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