No. of lines formed by joining 14 points =${}^{14}{C}_2$

No. of lines formed by joining 6 points =${}^6{C}_2$

Out of the 14 points, when you join the 6 collinear points, you don’t get${}^6{C}_2$lines, but you get only one line.

So, no. of lines${=}^{14}{C}_2{-}^6{C}_2+1=91-15+1=77$. 

2nd method:- By selection,  ${}^8{C}_2{+}^8{C}_1{\times }^6{C}_1+1=77$ .

A parallelogram is formed when two select 2 lines from the group of m lines, and 2 from the group of n line. This can be done in =${}^m{C}_2$and${}^n{C}_2$ways. So, total no. of parallelograms formed${=}^m{C}_2{\times }^n{C}_2$
$=\left(\frac{1}{4}\right)\left[\frac{(m!n!)}{\left(\right(m-2)!(n-2)!)}\right]$
$=\frac{1}{4}\left[mn\right(m-1\left)\right(n-1\left)\right]$ 

2nd method:- By substitution of any values in choices. Let $m=2\&n=2$ then total parallelograms =1. Only option C will satisfy. Then option (c).

Soln:

As n is odd,${}^n{C}_r$is maximum when r =$\frac{(n-1)}{2}$=15 or$\frac{(n-1)}{2}$= 16.

So, number of parties possible$={}^{31}{C}_{16}{=}^{31}{C}_{15}.$

Hence option (a)

We have 1 C, 1 H, 1 A, 1 I and 3 S.

The four letter word can be of the form: abcd, aabc, aaab

Case 1: abcd (none repeating): 4 letters can be selected in${}^5{C}_4$ways. Can be arranged in $4!$ Ways.

Case 2: aabc (two repeating): a can only be S. 2 can be selected in${}^4{C}_2$ways. Can be arranged in$\frac{4!}{(2!)}$ways.

Case 3: aaab (three repeating): a can only be S. 1 can be selected in${}^4{C}_1$ways. Can be arranged in$\frac{4!}{(3!)}$Ways.

So, total no. of words possible
$={(}^5{C}_4\times 4!)+{(}^4{C}_2\times \left(\frac{4!}{2!}\right)+{(}^4{C}_1\times \left(\frac{4!}{3!}\right)))$
$=120+72+16=208.$

The no. of ways in which mn different things can be divided equally into m groups containing n things each =$\left[\frac{(mn!)}{(n!{)}^m}\right]\left(\frac{1}{m!}\right)$

So, required no. of ways = $\left[\frac{(15!)}{(5!{)}^3}\right]\left(\frac{1}{3!}\right)$

We divide by$3!$ Since the groups are of equal size and hence, indistinguishable.

In distribution order is important.

The no. of ways in which mn different things can be distributed equally into m groups containing n things each =$\left[\frac{(mn!)}{(n!{)}^m}\right]$

Here, the groups are different, hence we do not divide by$3!.$

So, required no. of ways =$\left[\frac{(15!)}{(5!{)}^3}\right]$

No. of ways = $\frac{10!}{(6!4!)}$

2nd method:- Select any six then other four will be automatically selected. i.e${}^{10}{C}_6=\frac{10!}{(6!4!)}$

Here as it has to be divided among 4 different players, the order is important.

No. of ways =$\frac{(52!)}{\left[\right(13!{)}^4]}$. 

No. of ways =$\frac{10!}{(6!4!)}\times 2!$

Here we multiply by$2!$ Because the groups are distinct.

This is a type of “similar to different” questions with a lower limit of 1.

Let the 4 boxes be A,B,C,D

A+B+C+D=8

Applying a lower limit of 1

A+B+C+D=4

Applying 0’s and 1’s method:

No. of zeroes = 4 , No. of ones = 3

Total number of ways =$\frac{7!}{(4!3!)}$.