A committee of 6 is chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. If 2 particular women refuse to serve on the same committee, the number of ways of forming the committee is

$$\begin{gathered} (\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{selecting} 6\mathrm{members} \mathrm{with} \mathrm{atleast} 3 \mathrm{men},2 \mathrm{ladies}) \\ -(\mathrm{No} \mathrm{of} \mathrm{ways} of \mathrm{selecting} 6 \mathrm{members} \mathrm{in} \mathrm{which} \mathrm{both} \mathrm{specified} \mathrm{ladies} \mathrm{are} \mathrm{selected}) \\ \Rightarrow \left(10{\mathrm{c}}_3.7{\mathrm{c}}_3+10{\mathrm{c}}_4.7{\mathrm{c}}_2\right)-\left(10{\mathrm{c}}_3.5{\mathrm{c}}_1+10{\mathrm{c}}_4\cdot 5{\mathrm{c}}_0\right) \\ \Rightarrow \left(120\left(35\right)+\left(210\right)\left(21\right)\right)-\left(\left(120\right)5+210\right) \\ \Rightarrow \left(4200+4410\right)-\left(600+210\right) \\ \Rightarrow 8610-810 \\ \Rightarrow 7800 \end{gathered}$$