Question

I : Out of $7$ men and $4$ women a committee of $5$ is to be formed. The number of ways in which this can be done so as to include exactly $2$ women is $210$.
II : So as to include at least $2$ women is $301$. which of the above statements is true?

• A. only I
• B. only II
• C. Both I and II
• D. Neither I nor II

Answer 1

Answer: (C)

Both I and II

(I) Exactly $2$ women can be selected out of $4$ women in ${}^4{C}_2$ ways.

i.e., $\frac{4!}{(4-2)!2!}$

$=\frac{4\times 3\times 2!}{2!\times 2!}=6$ ways.

and remaining $3$ men can be selected out of $7$ men in ${}^7{C}_3$ ways.

i.e. $\frac{7!}{(7-3)!3!}=\frac{7!}{4!3!}=\frac{7\times 6\times 5}{6}=35$ ways.

No. of ways in which exactly 2 women are there in the group of $5$ is $6\times 35=210$

(II)No. of ways in which 2 women can be selected out of $4$ is $6$ ways(same as above)

No. of ways in which $3$ men can be selected is $35$ ways (as above).

No. of ways in which exactly $2$ women are there in the group of 5 is $=6\times 35=210$

No. of ways in which 3 women can be selected out of 4 is ${}^4{C}_3=4$ ways and remaining 2 men can be selected out of 7 men in ${}^7{C}_2=21$ ways.

No. of ways in which 3 women are there in the group of 5 is $=4\times 21=84$

No. of ways in which 4 women can be selected out of 4 is ${}^4{C}_4=1$ way and the remaining 1 men can be selected out of 7 men in ${}^7{C}_1=7$ ways.

No. of ways in which 4 women are there in the group of 5 is $1\times 7=7$

Total number of ways=$210+84+7=301$