(i) Since any individual can stand in any position, 8 women and 6 men can be arrange in 14P₁₄ = 14! ways
(ii) Considering 6 men as one unit, we have 9 people and they can be arranged in 9! ways. These 6 men can arrange among themselves in 6! ways.
∴ Total number of arrangement = \(9!\times 6!\)
(iii) 8 women can be arranged in a 8 places in 8! ways
\(\times \boxed { W } \times \boxed { W } \times \boxed { W } \times \boxed { W } \times \boxed { W } \times \boxed { W } \times \boxed { W } \times \boxed { W } \)
There are 9 (marked) places for 6 men
They can be arranged in 9P₆ ways
∴ Total number of ways \(9{ P }_{ 6 }\times 8!\)

Let p(n) be the statement
\({1\over 2.5}+{1\over 5.8}+{1\over 8.11}+...+{1\over {(3n-1)(3n+2)}}={n\over 6n+4}\)
Step 1: Putting n = 1, we get
\({1\over 2.5}={1\over 6(1)+4}⇒{1\over 10}={1\over 10}\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
\(∵\ {1\over 2.5}+{1\over 5.8}+...+{1\over (3K-1)(3K+2)}={K\over 6K+4}\)
Step. 3: To show that p(K + 1) is true
ie to P.T \({1\over 2.5+}+{1\over 5.8}+...+{1\over{(3K-1)(3K+2)}}={K\over 6+4}\)
ie to P.T \({1\over 2.5}+{1\over 5.8}+...+{1\over (3K-1)(3K+2)}+{1\over (3K+2)(3K+5)}={K+1\over 6K+10}\)
LHS = \({1\over 2.5}+{1\over 5.8}+...+{1\over (3k-1)(3k+2)}+{1\over (3k+2)(3k+5)}\)
\(={k\over 6k+4}+{1\over 5.8}+...+{1\over (3k-1)(3k+2)}+{1\over (3K+2)(3k+5)}\)
\(={k\over 6k+4}+{1\over (3k+2)(3k+5)}\) [using (1)]
\(={k\over 2(3k+2)}+{1\over (3k+2)(3k+5)}\)
\(={1\over 3k+2}\left[{k\over 2}+{1\over 3k+5}\right]\)
\(={1\over 3k+2}\left[k(3k+5)+2\over 2(3k+5)\right]={1\over 3k+2}\left[3k^2+5k+2\over 2(3k+5)\right]\)
\(={1\over 2(3k+5)(3k+5)}(k+1)(3k+2)\)
\(={k+1\over 2(3k+5)}={k+1\over 6k+10}=RHS\)

∴ p (k + 1) is line.
Hence, by the principle mathematical induction p(n) is true for all values of n.

Adding 1 both sides to the given statement
p(n): \(1+{1\over 1+2}+{1\over 1+2+3}+..+{1\over 1+2+3+.+n}=1+{n-1\over n+1}={n+1+n-1\over n+1}={2n\over n+1}\)
Step 1:
Putting n = 2
\(1+{1\over 1+2}={2(2)\over 2+1}⇒1+{1\over 3}={4\over 3}⇒{4\over3}={4\over 3}\)
∵ p(1) is true
Step 2:
Let us assume that p(K) is true
\(∵\ 1+{1\over 1+2}+...+{1\over 1+2+3+...+K}={2K\over K+1}\)
Step 3:
To prove that p(K+1) is true
ie \(1+{1\over 2}+...+{1\over 1+2+3+K}+{1\over 1+2+...+K+1}={2(K+1)\over K+2}\)
LHS = \(1+{1\over 1+2}+...+{1\over 1+2+3+...+K}+{1\over 1+2+...+(K+1)}\)
\(={2K\over K+1}+{1\over 1+2+3+..+(K+1)}\) [using (1)]
\(={2K\over K+1}+{1\over {(K+1)(K+2)\over2}}\) \(\left[ ∵\ \sum n={n(n+1)\over2} \right]\)
\(={2K\over K+1}+{2\over (K+1)(K+2)}={2\over K+1}\left[ K+{1\over K+2}\right]={2\over K+1}\left(K^2+2K+1\over K+2\right)\)
\(={2(K+1)^2\over (K+1)(K+2)}={2(K+1)\over K+2}=RHS\)
∵ p(K+1) is true
Hence, by mathematical induction, p(n) is true for all values of n

Let p(n) be the statement
\(1^3+2^3+3^3+...+n^3=\left[{n(n+1)\over2}\right]\)
Step1: Put n = 1
\(⇒\ 1^3=\left[1(1+1)\over 2\right]^2=\left[1(2)\over2\right]^2\)
⇒ 1 = 1
∴ p(n) is true
Step 2: Let us assume that p(K) is true
∵ \(1^3 + 2^3 + 3^3 + ... +K^3​​​​​​​ =​​​​​​​\left[K(K+1)\over 2\right]^2\)
Step 3: To prove that p(K+1) is true
i.e to P.T \(1^3+2^3+...+K^3+(K+1)^3=\left[(K+1)(K+2)\over 2\right]^2\)
LHS = 1³ + 2³ + .... + K³ + (K + 1)³
\(=\left[K{(K+1)\over 2}\right]^2+(K+1)^3\)
\(=(K+1)^2\left[{K^2\over 4}+K+1\right]=(K+1)^2\left[K^2+4K+4\over 4\right]\)
\(={(K+1)^2(K+2)^2\over4}=\left[(k+1)(k+2)\over 2\right]=RHS\)
∵ p(K+1) is true
∵ By the principle of mathematical induction, p(n) is true for all values of n.

(i) Either starts with L or ends with S.

1	4	3	2	1
L

Since the words starts with L, the remaining 4 boxes can be filled in 4 x 3 x 2 x 1 ways by the remaining letters 0, T, U, S.
∴ Number of words starting with L
= 1 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24.

1	2	3	4	1
				S

Here also, the remaining 4 boxes can be filled in 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 ways = 24.......(1)
Number of words ending with S = 24 ....(2)
Number of words starting with L and end with S are 3 \(\times\) 2 \(\times\) 1 = 6...(3)
∴  By fundamental principle of addition, number of words either starts with L nor ends with S = 24 + 24 - 6 = 48 - 6 = 42

	3	2	1
F				S

(ii) Neither starts with L nor ends with S.
Total number of words formed by the letters of the word LOTUS is 5 \(\times\) 4 \(\times\) 3 \(\times\) 2\(\times\) 1 = 120.
Now, number of words neither starts with L nor end with S.
= (Total number of words) - (Number of words starts with either L nor ends with S)
= 120 - 42
= 78.

The given data can be converted as a route map diagram as follows:

The possible choices for number of routes commuting from A to place C via place b, without using similar mode of transportation are
( B₁, T'₁) ( B₂, T'₂ ) ( B₁, A₁ ) ( B₂, T'₁ ) ( B₂, T'₂ ) ( B₂, A'₁ ) ( T₁, A'₁ ) ( T₂, B'₁ ) ( T₂, A'₁ ) ( A_1, B'₁ ) ( A_1, T'₁ )  and ( A_2, T'₂ ) 
Required number of routes of commuting from place A to place C via place 'B'
\(=(2 \times 3)+(2 \times 2)+(1 \times 3)\)
= 6+4 + 0 = 7
Required number of routes  = 20 - 7 = 13

(i) No restriction.
Given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Since we need numbers between 999 and 10000, it has only 4 digits.

thousands	hundreds	tens	ones
9	10	10	10

Thousands place can be filled in 9 ways (excluding 0) since there is no restriction, hundreds, place, tens place and unit place can be filled in 10 ways each using all the digits.
∴ By fundamental principle of multiplication, 9 required number of 4 - digit numbers.
= 9 \(\times\) 10 \(\times\) 10 \(\times\) 10
= 9000.
(ii) No digit is repeated.

thousands	hundreds	tens	ones
9	10	10	10

Thousands place can be filled in 9 ways (excluding 0)
Since repetition is not allowed, unit place can be filled in 9 ways tens place can be filled in 8 ways and hundreds place can be filled in 7 ways.
∴ By fundamental principle of multiplication, required number of 4 digit numbers
= 9 \(\times\) 7 \(\times\) 8 \(\times\) 9
= 4536.
(iii) At least one of the digits is repeated.
Required number of numbers = Total number of 4 digit numbers - number of 4 digit numbers when no digit is repeated
= 900 - 4536 = 4464

The repetition of digits is allowed

Hundreds	tens	unit
5	6	3

The unit place can be filled in 3 ways using the digits 1, 3, or 5 since we need 3 digit odd numeric Hundreds place can be filled in 5 ways excluding 0 and repetition of digits is allowed.
Tens place can be filled in 6 ways .
∴ By fundamental principle of multiplication, required number of 3 = digit odd numbers
= 5 \(\times\) 6 \(\times\) 3 = 30 \(\times\) 3 = 90.

The Repetition of digits is not allowed

Hundreds	tens	unit
4	4	3

Since we need 3 - digit odd numbers, the unit place can be filled in 3 ways using the digits 1, 3 or 5.
Hundred's place can be filled in 4 ways (excluding 0 and the number used for one's place) since repetition is not allowed.
Ten's place can be filled 4 ways including 0.
∴ By fundamental principle of multiplication, number of required 3 = digit odd numbers
= 4 \(\times\) 4 \(\times\) 3 = 48

(i) Repetition of digit is allowed

4

6

6

Since we are going to find numbers between 100 and 500 it has 3 = digits
The unit place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5
The tens place also can be filled in 6 ways since repetition of digits is allowed.
The hundreds place can be filled in 4 ways using the digits 1, 2, 3, 4 [excluding 0 and 5]
∴ By fundamental principle of multiplication, required number of 3 - digit numbers = 4 \(\times\) 6 \(\times\) 6 = 144.
(ii) Repetition of digits is not allowed.

4

5

4

Hundreds place can be filled in 4 ways excluding 0 and 5
Tens place can be filled in 5 ways since repetition of digits is not allowed
Unit place can be filled in 4 ways.
∴ By fundamental principle of multiplication, required number of three-digit numbers = 4 \(\times\) 5 \(\times\) 4 = 80.