If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
It is given that no two lines are parallel means all line are intersecting and no three lines are concurrent means three lines intersect at a point.
Since, we know that for one point of intersection, we required two lines.
$$\begin{aligned} \therefore \text { Number of point of intersection } & ={ }^{20} C_2=\frac{20!}{2!18!}=\frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} \\ & =\frac{20 \times 19}{2}=19 \times 10=190 \end{aligned}$$
In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64 . How many telephone numbers have all six digits distinct?
If first two digit is 41 , the remaining 4 digits can be arranged in
$$\begin{aligned} & ={ }^8 P_4=\frac{8!}{8-4!}=\frac{8!}{4!} \\ & =\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!} \\ & =8 \times 7 \times 6 \times 5=1680 \end{aligned}$$
Similarly, if first two digit is $42,46,62$, or 64 , the remaining 4 digits can be arranged in ${ }^8 P_4$ ways i.e., 1680 ways.
$\therefore$ Total number of telephone numbers have all six digits distinct $=5 \times 1680=8400$
In an examination, a student has to answer 4 questions out of 5 questions, questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
It is given that 2 questions are compulsory out of 5 questions.
So, these two questions are always included in the selection.
We know that, the selection of $n$ distinct objects taken $r$ at a time in which $p$ objects are always included in ${ }^{n-p} C_{r-p}$ ways.
$$\begin{aligned} \therefore \text { Total number of ways } & ={ }^{5-2} C_{4-2}={ }^3 C_2 \\ & =\frac{3!}{2!1!}=\frac{3 \times 2!}{2!}=3 \end{aligned}$$
If a convex polygon has 44 diagonals, then find the number of its sides.
Let the convex polygon has $n$ sides.
$\therefore$ Number of diagonals $={ }^n C_2-n$
According to the question,
$$\begin{aligned} & { }^n C_2-n=44 \\ & \frac{n!}{2!(n-2)!}-n=44 \\ & \Rightarrow \quad \frac{n(n-1)}{2}-n=44 \\ & \Rightarrow \quad n\left[\frac{(n-1)}{2}-1\right]=44 \quad \Rightarrow n\left(\frac{n-1-2}{2}\right)=44 \\ & \Rightarrow \quad n(n-3)=44 \times 2 \Rightarrow n(n-3)=88 \\ & \Rightarrow \quad n^2-3 n-88=0 \quad \Rightarrow(n-11)(n+8)=0 \\ & \Rightarrow \quad n=11,-8 \\ & \therefore \quad n=11\quad [\because n \neq-8] \end{aligned}$$
18 mice were placed in two experimental groups and one control group with all groups equally large. In how many ways can the mice be placed into three groups?
It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
$\therefore \quad$ Required arrangements $=\frac{\text { Total arrangement }}{\text { Equally likely arrangement }}=\frac{18!}{6!6!6!}$