Question

A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. The number of different ways this can be done, if two particular women refuse to serve on the same committee is ___________.

Answer 1

Here, Number of men = 10
Number of women = 7
6 committee numbers can be selected containing at-least 3 men and 2 women in following 2 ways.
→ 4 men and 2 women
→ 3 men and 3 women
∴ number of ways of selecting at-least 3 men and 2 women in committee of 6
$$\begin{gathered} {=}^{10}{C}_3{\times }^7{C}_3{+}^{10}{C}_4{+}^7{C}_2 \\ =\frac{10\times 9\times 7}{3\times 2}\times \frac{7\times 6\times 5}{6}+\frac{10\times 9\times 8\times 7}{4\times 3\times 2}\times \frac{7\times 6}{2} \\ =120\times 35+210\times 21 \\ =8610 \end{gathered}$$
Number of ways when two particular women are never together = ¹⁰C₃
$$\begin{gathered} =\frac{{}^{10}C_4}{\mathrm{all}\mathrm{men}}+\frac{{}^{10}C_3\times {}^{5}C_1}{2\mathrm{women}} \\ =210+120\times 5 \\ =210+600 \\ =810 \end{gathered}$$
Therefore, Total number of ways when two particular women are never together = 8610 − 810 = 7800.