A committee of 6 is to be chosen from 10 men and 7 women, so as to contain atleast 3 men and 2 women. In how many different ways can this be done, if two particular women refuse to serve on the same committee?
$\because\quad$ Total number of men = 10
and total number of women =7
We have to form a committee containing atleast 3 men and 2 women.
Number of ways $={ }^{10} \mathrm{C}_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2$
If two particular women to be always there .
$\therefore \quad$ Number of ways $={ }^{10} C_4 \times{ }^5 C_0+{ }^{10} C_3 \times{ }^5 C_1$
Total number of committee when two particular women are never together
$$\begin{aligned} & =\text { Total }- \text { Together } \\ & =\left({ }^{10} C_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2\right)-\left({ }^{10} C_4 \times{ }^5 C_0+{ }^{10} C_3 \times{ }^5 C_1\right) \\ & =(120 \times 35+210 \times 21)-(210+120 \times 5) \\ & =4200+4410-(210+600) \\ & =8610-810=7800 \end{aligned}$$
If ${ }^n C_{12}={ }^n C_8$, then $n$ is equal to
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