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 $=\frac{\text { Total arrangement }}{\text { Equally likely arrangement }}=\frac{18!}{6!6!6!}$
A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if (i) they can be of any colour. (ii) two must be white and two red. (iii) they must all be of the same colour.
Total number of marbles $=6$ white +5 red $=11$ marbles
(i) If they can be of any colour means we have to select 4 marbles out of 11.
$\therefore$ Required number of ways $={ }^{11} C_4$
(ii) If two must be white, then selection will be ${ }^6 \mathrm{C}_2$ and two must be red, then selection will be ${ }^5 \mathrm{C}_2$.
$\therefore$ Required number of ways $={ }^6 C_2 \times{ }^5 C_2$
(iii) If they all must be of same colour, then selection of 4 white marbles out of $6={ }^6 C_4$
and selection of 4 red marble out of $5={ }^5 C_4$
$\therefore \quad$ Required number of ways $={ }^6 C_4+{ }^5 C_4$