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 $={ }^6 C_4+{ }^7 C_6$
$$=15+7=22$$
How many committee of five person with a chairperson can be selected from 12 persons?
$\because \quad$ Total number of persons $=12$
and number of persons to be selected $=5$
Out of 12 persons a chairperson is selected $={ }^{12} C_1=12$ ways
Now, remaining 4 persons are selected out of 11 persons.
$\therefore \quad$ Number of ways $={ }^{11} C_4=330$
$\therefore$ Total number of ways to form a committee of 5 persons $=12 \times 330=3960$
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
There are 26 English alphabets and 10 digits (0 to 9).
Since, it is given that each plate contains two different letters followed by three different digits.
$\therefore$ Arrangement of 26 letters, taken 2 at a time $={ }^{26} P_2=\frac{26!}{24!}=26 \times 25=650$
and three-digit number can be formed out of the 10 digits $={ }^{10} P_3=10 \times 9 \times 8=720$ ways
$\therefore$ Total number of licence plates $=650 \times 720=468000$
A bag contains 5 black and 6 red balls, determine the number of ways in which 2 black and 3 red balls can be selected from the lot.
It is given that bag contains 5 black and 6 red balls.
So, 2 black balls is selected from 5 black balls in ${ }^5 \mathrm{C}_2$ ways.
and 3 red balls are selected from 6 red balls in ${ }^6 C_3$ ways.
$\therefore$ Total number of ways in which 2 black and 3 red balls are selected $={ }^5 C_2 \times{ }^6 C_3$
$$=10 \times 20=200 \text { ways }$$