$$\begin{aligned} \text { Total number of ways } & ={ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2+{ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_6+\ldots+{ }^{10} \mathrm{C}_{10} \\ & =2^{10}-1 \quad [\because \quad ^nC_0+^nC_1+^nC_2+...=2^n]\\ & =1024-1=1023 \end{aligned}$$

There are 2 white, three black and four red balls.

We have to draw 3 balls, out of these 9 balls in which atleast one black ball is included. Hence, we can select the balls in the following ways.

$$\begin{aligned} \therefore \text { Required number of selections } & ={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2+{ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1+{ }^3 \mathrm{C}_3 \times{ }^6 \mathrm{C}_0 \\ & =3 \times 15+3 \times 6+1 \\ & =45+18+1=64 \end{aligned}$$

$$\begin{array}{lr} \text { Given, } & { }^n C_{r-1}=36 \quad \text{... (i)}\\ \Rightarrow & { }^n C_r=84 \quad \text{... (ii)}\\ \Rightarrow & { }^n C_{r+1}=126 \quad \text{... (iii)} \end{array}$$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{36}{84} \quad\left[\begin{array}{l}\because{ }^n C_r=\frac{n!}{(n-r!r!)} \\ \text { and } n!=n(n-1)!\end{array}\right]$

$$\begin{array}{ll} \Rightarrow & \frac{n!}{(r-1)!\{n-(r-1)\}!} \cdot \frac{r!(n-r)!}{n!}=\frac{3}{7} \\ \Rightarrow & \frac{1}{(r-1)!(n-r+1)!} \cdot \frac{r(r-1)!(n-r)!}{1}=\frac{3}{7} \\ \Rightarrow & \frac{1 \cdot r}{(n-r+1)(n-r)!} \cdot(n-r)!=\frac{3}{7} \Rightarrow \frac{r}{n-r+1}=\frac{3}{7} \\ \Rightarrow & 7 r=3 n-3 r+3 \\ \Rightarrow & 10 r-3 n=3\quad \text{(iv)} \end{array}$$

On dividing Eq. (ii) by Eq. (iii), we get

$$\begin{array}{rlrl} & \frac{{ }^n C_r}{{ }^n C_{r+1}} =\frac{84}{126} \\ \Rightarrow & \frac{n!}{r!(n-r)!} \cdot \frac{(r+1)!(n-r-1)!}{n!} =\frac{14}{21} \\ \Rightarrow & \frac{1}{r!(n-r)!(n-r-1)!} \cdot \frac{(r+1) r!(n-r-1)!}{r} =\frac{2}{3} \Rightarrow \frac{r+1}{n-r}=\frac{2}{3} \\ \Rightarrow & 3 r+3=2 n-2 r \Rightarrow \quad 2 n-5 r=3\quad \text{... (v)} \end{array}$$

On multiplying Eq. (iv) by 2 and Eq. (v) by 3, we get

$$\begin{aligned} & 20 r-6 n=6 \quad \text{... (vi)}\\ & 6 n-15 r=9\quad \text{... (vii)} \end{aligned}$$

On adding Eqs. (vi) and (vii),

From Eq. (v),

$$\begin{aligned} 5 r & =15 \quad \Rightarrow \quad r=3 \\ 2 n & =3+15 \\ 2 n & =18 \quad \Rightarrow \quad n=9 \end{aligned}$$

$$\therefore \quad \quad{ }^r C_2={ }^3 C_2=\frac{3!}{2!1!} \quad=\frac{3 \times 2!}{2!}=3$$

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.

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.

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