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