Here, we have to find the number of integers greater than 7000 with the digits $3,5,7,8$ and 9. So, with these digits we can make maximum five-digit number because repeatition is not allowed.

Now, all the five-digit numbers are greater than 7000 .

Number of ways of forming 5 -digit number $=5 \times 4 \times 3 \times 2 \times 1=120$

and all the four-digit numbers greater than 7000 can be formed in following manner.

Thousand place can be filled in 3 ways. Hundred place can be filled in 4 ways. Tenth place can be filled in 3 ways. Units place can be filled in 2 ways.

Thus, we have total number of 4-digit number $=3 \times 4 \times 3 \times 2=72$

$\therefore$ Total number of integers $=120+72=192$

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.

$$\begin{aligned} \therefore \text { Number of point of intersection } & ={ }^{20} C_2=\frac{20!}{2!18!}=\frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} \\ & =\frac{20 \times 19}{2}=19 \times 10=190 \end{aligned}$$

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 $=5 \times 1680=8400$

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.

$$\begin{aligned} \therefore \text { Total number of ways } & ={ }^{5-2} C_{4-2}={ }^3 C_2 \\ & =\frac{3!}{2!1!}=\frac{3 \times 2!}{2!}=3 \end{aligned}$$

Let the convex polygon has $n$ sides.

$\therefore$ Number of diagonals $={ }^n C_2-n$

According to the question,

$$\begin{aligned} & { }^n C_2-n=44 \\ & \frac{n!}{2!(n-2)!}-n=44 \\ & \Rightarrow \quad \frac{n(n-1)}{2}-n=44 \\ & \Rightarrow \quad n\left[\frac{(n-1)}{2}-1\right]=44 \quad \Rightarrow n\left(\frac{n-1-2}{2}\right)=44 \\ & \Rightarrow \quad n(n-3)=44 \times 2 \Rightarrow n(n-3)=88 \\ & \Rightarrow \quad n^2-3 n-88=0 \quad \Rightarrow(n-11)(n+8)=0 \\ & \Rightarrow \quad n=11,-8 \\ & \therefore \quad n=11\quad [\because n \neq-8] \end{aligned}$$