Question

A committee of 7 has to be formed from 9 boys and 4 girls.
Statement I The number of ways this can be done, if committee consists of exactly 3 girls equals $504$.
Statement II The number of ways this can be done, if committee consists of atleast 3 girls equal $588$.
Statement III The number of ways this can be done, if committee consists of atmost 3 girls equal $1632$.
Identify the correct combination of true (T) and false $(F)$ of the given three statements

• A. TTT
• B. TTF
• C. TFF
• D. FTT

Answer 1

Answer: (A)

I. Out of 7 persons, we want to select exactly 3 girls i.e., remaining 4 person will be boys.

4 boys out of 9 boys can be selected in ${}^9{C}_4$ ways.
3 girls out of 4 girls can be selected in ${}^4{C}_3$ ways.
Hence, by FPC, total number of ways for making the committee
$={}^9{C}_4\times {}^4{C}_3$
$=\frac{9\times 8\times 7\times 6}{24}\times {}^4{C}_1\left(\because {}^n{C}_r={}^n{C}_{n-r}\right)$
$=\frac{9\times 8\times 7\times 6\times 1}{24}\left(\because {}^n{C}_1=n\right)$
$=9\times 8\times 7\left[\because {}^n{C}_4=\frac{n(n-1)(n-2)(n-3)}{24}\right]$
$=72\times 7=504$ ways
II. Here, we have to select atleast 3 girls.
There are following two possibilities

In first case, 4 boys out of 9 boys can be selected in ${}^9{C}_4$ ways and 3 girls out of 4 girls can be selected in ${}^4{C}_3$ ways.
Hence, by FPC, total number of ways in this case
$={}^9{C}_4\times {}^4{C}_3=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}\times 4$
$=504$ ways
In second case, 3 boys out of 9 boys can be select in ${}^9{C}_3$ ways and 4 girls out of 4 girls can be select in ${}^4{C}_4$ ways.
Hence, by FPC, total number of ways in this case
$={}^9{C}_3\times {}^4{C}_4$
$=\frac{9\times 8\times 7}{6}\times 1=84$ ways
Hence, by fundamental principle of addition, total number of ways $=504+84=588$
III. Here, we have to select atmost 3 girls i.e., there are following four possibilities

Hence, total number of ways (as these events are dependent to each other)
$=\left({}^9{C}_4\times {}^4{C}_3\right)+\left({}^9{C}_5\times {}^4{C}_2\right)+\left({}^9{C}_6\times {}^4{C}_1\right)+\left({}^9{C}_7\times {}^4{C}_0\right)$
$=\left({}^9{C}_4\times {}^4{C}_1\right)+\left({}^9{C}_4\times {}^4{C}_2\right)+\left({}^9{C}_3\times {}^4{C}_1\right)+\left({}^9{C}_2\times {}^4{C}_0\right)$
$\left(\because {}^n{C}_r={}^n{C}_{n-r}\right)$
$=\left(\frac{9\times 8\times 7\times 6}{24}\times 4\right)+\left(\frac{9\times 8\times 7\times 6}{24}\times \frac{4\times 3}{2}\right)$
$+\left(\frac{9\times 8\times 7}{6}\times 4\right)+\left(\frac{9\times 8}{2}\times 1\right)$
$=504+756+336+36$
$=1632$ ways