Question

From a group of $6$ men and $7$ women, five persons are to be selected from a committee so that at least $3$ women are there on the committee. In how any ways can it be done?

• A. $576$
• B. $756$
• C. $657$
• D. $765$

Answer 1

The correct option is B $756$

Since atleast $3$ women must, be chosen we consider all committees which include $3,4,5$ women with $2,1,0$ men respectively.

a) Choose $3$ from $7$ women and $2$ from $6$ men $={}^7{C}_3\ast {}^6{C}_2=35\times 15=525$

b) Choose $4$ from $7$ women and $1$ from $6$ men $={}^7{C}_4\times {}^6{C}_1=35\ast 6=210$.

c) Choose $5$ from $7$ women and $0$ from $6$ men $={}^7{C}_5\times {}^6{C}_0=21\times 1=21$

$\therefore a+b+c=525+210+21=756$.