In how many ways can a football team of 11 players be selected from 16 players? How many of them will
(i) include 2 particular players?
(ii) exclude 2 particular players?
Total number of players $=16$
We have to select a team of 11 players
(i) include 2 particular players $={ }^{16-2} C_{11-2}={ }^{14} C_9$
[since, selection of $n$ objects taken $r$ at a time in which pobjects are always included is ${ }^{n-p} C_{r-p}$ ]
(ii) Exclude 2 particular players $={ }^{16-2} C_{11}={ }^{14} C_{11}$
[since, selection of $n$ objects taken $r$ at a time in which p objects are never included is $\left.{ }^{n-p} C_r\right]$
A sports team of 11 students is to be constituted, choosing atleast 5 from class XI and atleast 5 from class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Total students in each class $=20$
We have to selects atleast 5 students from each class.
Hence, selection of sport team of 11 students from each class is given in following table
| Class XI | 5 | 6 |
|---|---|---|
| Class XII | 6 | 5 |
$\begin{aligned} \therefore \text { Total number of ways of selecting a team of } 11 \text { players } & ={ }^{20} C_5 \times{ }^{20} C_6+{ }^{20} C_6 \times{ }^{20} C_5 \\ & =2 \times{ }^{20} C_5 \times{ }^{20} C_6\end{aligned}$
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected, if the team has
(i) no girls.
(ii) atleast one boy and one girl.
(iii) atleast three girls.
Number of girls $=4$ and Number of boys $=7$
We have to select a team of 5 members provided that
(i) team having no girls.
$$\therefore \quad \text { Required selection }={ }^7 C_5=\frac{7!}{5!2!}=\frac{7 \times 6}{2}=21$$
(ii) atleast one boy and one girl
$$\begin{aligned} \therefore \text { Required selection } & ={ }^7 C_1 \times{ }^4 C_4+{ }^7 C_2 \times{ }^4 C_3+{ }^7 C_3 \times{ }^4 C_2+{ }^7 C_4 \times{ }^4 C_1 \\ & =7 \times 1+21 \times 4+35 \times 6+35 \times 4 \\ & =7+84+210+140=441 \end{aligned}$$
(iii) when atleast three girls are included $={ }^4 C_3 \times{ }^7 C_2+{ }^4 C_4 \times{ }^7 C_1$
$$=4 \times 21+7=84+7=91$$
A committee of 6 is to be chosen from 10 men and 7 women, so as to contain atleast 3 men and 2 women. In how many different ways can this be done, if two particular women refuse to serve on the same committee?
$\because\quad$ Total number of men = 10
and total number of women =7
We have to form a committee containing atleast 3 men and 2 women.
Number of ways $={ }^{10} \mathrm{C}_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2$
If two particular women to be always there .
$\therefore \quad$ Number of ways $={ }^{10} C_4 \times{ }^5 C_0+{ }^{10} C_3 \times{ }^5 C_1$
Total number of committee when two particular women are never together
$$\begin{aligned} & =\text { Total }- \text { Together } \\ & =\left({ }^{10} C_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2\right)-\left({ }^{10} C_4 \times{ }^5 C_0+{ }^{10} C_3 \times{ }^5 C_1\right) \\ & =(120 \times 35+210 \times 21)-(210+120 \times 5) \\ & =4200+4410-(210+600) \\ & =8610-810=7800 \end{aligned}$$
If ${ }^n C_{12}={ }^n C_8$, then $n$ is equal to