Eight chairs are numbered 1 to 8 . Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
First women choose the chairs from among 1 to 4 chairs. i.e., total number of chairs is 4 .
Since, there are two women, so number of arrangements $={ }^4 P_2$ ways.
Now, men have to choose chairs from remaining 6 chairs.
Since, there are 3 men, so number can be arranged in ${ }^6 P_3$ ways.
$$\begin{aligned} \therefore \text { Total number of possible arrangements } & ={ }^4 P_2 \times{ }^6 P_3 \\ & =\frac{4!}{4-2!} \times \frac{6!}{6-3!} \\ & =\frac{4!}{2!} \times \frac{6!}{3!} \\ & =\frac{4 \times 3 \times 2!}{2!} \times \frac{6 \times 5 \times 4 \times 3!}{3!} \\ & =4 \times 3 \times 6 \times 5 \times 4=1440 \end{aligned}$$
If the letters of the word 'RACHIT' are arranged in all possible ways as listed in dictionary. Then, what is the rank of the word 'RACHIT'?
The letters of the word 'RACHIT' in alphabetical order are A, C, H, I, R and T.
Now, words beginning with $A=5$ !
words beginning with $C=5$ !
words beginning with $H=5$ !
words beginning with $I=5$ !
Word beginning with R i.e., RACHIT $=1$
$\therefore \quad$ Rank of the word 'RACHIT' in dictionary $=4 \times 5!+1=4 \times 120+1$
$=480+1=481$
A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.
Since, candidate cannot attempt more than 5 questions from either group. Thus, he is able to attempt minimum two questions from either group. The number of questions attempted from each group is given in following table
| Group I | 5 | 4 | 3 | 2 |
|---|---|---|---|---|
| Group II | 2 | 3 | 4 | 5 |
$$\begin{aligned} &\text { Since, each group have } 6 \text { questions and total attempted } 7 \text { questions. }\\ &\begin{aligned} \therefore \text { Total number of possible ways } & ={ }^6 C_5 \times{ }^6 C_2+{ }^6 C_4 \times{ }^6 C_3+{ }^6 C_3 \times{ }^6 C_4+{ }^6 C_2 \times{ }^6 C_5 \\ & =2\left[{ }^6 C_5 \times{ }^6 C_2+{ }^6 C_4 \times{ }^6 C_3\right] \\ & =2[6 \times 15+15 \times 20] \\ & =2[90+300] \\ & =2 \times 390=780 \end{aligned} \end{aligned}$$
Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.
Total number of points $=18$
Out of which 5 points are collinear, we get a straight line by joining any two points.
$\therefore$ Number of straight line formed by joining the 18 points taking 2 at a time $={ }^{18} \mathrm{C}_2$ and number of straight line formed by joining 5 points taking 2 at a time $={ }^5 \mathrm{C}_2$
But 5 collinear points, when joined pairwise give only one line.
$$\begin{aligned} \therefore \text { Required number of straight line } & ={ }^{18} \mathrm{C}_2-{ }^5 \mathrm{C}_2+1 \\ & =153-10+1=144 \end{aligned}$$
We wish to select 6 person from 8 but, if the person $A$ is chosen, then $B$ must be chosen. In how many ways can selections be made?
Total number of person $=8$
Number of person to be selected $=6$
It is given that, if $A$ is chosen then, $B$ must be chosen.
Therefore, following cases arise.
Case I When $A$ is chosen, $B$ must be chosen.
Number of ways $={ }^{8-2} C_{6-2}={ }^6 C_4$
Case II When A is not chosen.
Then, $B$ may be chosen.
$\therefore \quad$ Number of ways $={ }^{8-1} C_6={ }^7 C_6$
Hence, required number of ways $={ }^6 C_4+{ }^7 C_6$
$$=15+7=22$$