Question

A commitee of six is chosen from ten men and seven women so as to contain atleast three men and two women. If two particular women refuse to serve on the same committee, the number of ways of forming the committee is:

• A. $7800$
• B. $8610$
• C. $810$
• D. $8000$

Answer 1

The correct option is A $7800$

Case $1:$ Let the commitee has $2$ women, then total number of ways commitee can be formed without any restriction $={}^{10}{C}_4\cdot {}^7{C}_2=4410$

Case $2:$ Let the commitee has $3$ women, then total number of ways commitee can be formed without any restriction $={}^{10}{C}_3\cdot {}^7{C}_3=4200$

Now, if two particular womens are in same commitee, then total number of ways commitee can be formed
$={}^{10}{C}_3\cdot {}^5{C}_1+{}^{10}{C}_4\cdot {}^5{C}_0=810$

$\therefore$ From complementary principle, required number of ways commitee can be formed
$=4410+4200-810=7800$