In the letters of the word, ARTICLE, there are three vowels namely A, I, E.
There are 3 even places.
3 vowels can occupy the even places in 3P₃ = 3! ways.
Remaining 4 letters can occupy 4 places in 4! ways.
Hence, total number of ways of arrangement = 4! \(\times\) 3!
=  \(4\times 3\times 2\times 3\times 2\)
=144

(i) With repetition

The given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The unit place can be filled in only one way using 3.
Since repetition is allowed, the tens place can be filled in 10 ways using any one of the digits from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The hundreds place can be filled in 9 ways using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (excluding 0)
∴ By fundamental principle of multiplication, total number of 3 digit numbers = 9 \(\times\) 10 \(\times\) 1 = 90.
(ii) Repetition of digits is not allowed

The unit place can be filled in 1 way
Since repetition of digits is not allowed, the tens place can be filled in 8 ways.
Hundreds place can be filled in 8 ways.
∴ Total number of 3-digit numbers without repetition = 1 \(\times\) 8 \(\times\) 8 = 64

(i)Repetition of digits is allowed

The unit place can be filled in 4 ways.
Since repetition is allowed, the tens place and hundreds place can also be filled in 4 ways each.
∴ Total number one-digit numbers = 4 x 4 x 4 = 64
(ii) Repetition of digits is not allowed

The unit place can be filled in 4 ways.
Since repetition of digits is not allowed, the tens place can be filled in 3 ways.
Hundreds place can be filled in 2 ways .
∴ Total number of 3-digit numbers without repetition = 4 \(\times\) 3 \(\times\) 2 = 24

(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 \(9{ P }_{ 6 }\times 8!\)

(i) Repetition of digit is allowed

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.

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.

\(1+3+5+7+\ldots \ldots+17 \text { is equal to } 9^{2}=81\)

\(\text { Given }{ }^{n} C_{4},{ }^{n} C_{5},{ }^{n} C_{6} \text { are in A.P }\)
\({ }^{2 n} \mathrm{C}_{5}={ }^{n} \mathrm{C}_{4}+{ }^{n} \mathrm{C}_{6}\)
\(\frac{2\lfloor n}{\lfloor n-5\lfloor 5}=\frac{\lfloor n}{\lfloor n -4\lfloor 4}+\frac{n}{\lfloor n -6\lfloor 6}\)
\(\frac{2}{\lfloor n-5\lfloor 5} =\frac{1}{\operatorname{\lfloor n}-4\lfloor 4}+\frac{1}{\lfloor n-6\lfloor 6}\)
\(\frac{2(n-4) 6}{(n-4)\lfloor n-5\lfloor 5.6}=\frac{5.6}{\lfloor-45.6\lfloor 4}+ \frac{(n-4)(n-5)}{\lfloor 6(n-4)(n-5) \lfloor n-6}\)
\(\Rightarrow \frac{12(n-4)}{\lfloor n-4\lfloor 6}=\frac{30}{\lfloor n-4\lfloor 6}+ \frac{(n-4)(n-5)}{\lfloor n-4 \lfloor 6}\)
\(12 n-48 =30+n^{2}-9 n+20 \)
\(n^{2}-21 n+98 =0 \)
\((n-14)(n-7) =0 \)
\(n=14(\text { or }) n =7 \)

\(\frac{{ }^{2 n} C_{3}}{{ }^{n} C_{3}} =\frac{11}{1} \)
\(\frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)} =\frac{11}{1} \)
\(\frac{2 n(2 n-1) 2(n-1)}{n(n-1)(n-2)} =11 \)
\(4(2 n-1) =11(n-2) \)
\(8 n-4 =11 n-22 \)
\(18 =3 n \)
\(n = 6\)

\(\text { Number of triangles }={ }^{10} \mathrm{C}_{3}-{ }^{4} \mathrm{C}_{3}\)
\(=\frac{10 \times 9 \times 8}{1 \times 2 \times 3}-4 \)
\(=120-4 =116 \)

\(\text { Number of points of intersection }={ }^{10} \mathrm{C}_{2}\)
\(=\frac{10 \times 9}{1 \times 2}=45\)

\(\text { Number of diagonals }={ }^{n} C_{2}-n\)
\(\frac{n(n-1)}{2}-n=44 \)
\(n^{2}-n-2 n=88 \)
\(n^{2}-3 n-88=0 \)
\((n-11)(n+8)=0 \)
\(n=11(\text { or }) n=-8 \)

\(a^{2}-a^{a} C_{2} =a^{2}-a^{a} C_{4} \)
\(a^{2}-a^{a} C_{2} =a^{2}-a^{a} C_{a-a-4}\left(\because^{n} C_{r}={ }^{n} C_{n-r}\right) \)
\(a^{2}-a-4 =2 \)
\(a^{2}-a-6 =0 \)
\((a-3)(a+2) =0 \)
\(a=3 \text { or } a=-2 \text { which is impossible }\)

The number of five digit telephone numbers which can be formed using the digits 0,1,2.... 9 is 10⁵.
The number of 5 digit numbers which has none of their digit repeated is ¹⁰P₅, = 30240
The required number of telephone. number is 10⁵ =- 30240 = 69,760

Product of r consecutive positive integers is divisible by r! (by theorem).

\({ }^{(n+5)} P_{(n+1)} \quad=\frac{11(n-1)}{2}{ }^{n+3} P_{n}\)
\(\frac{(n+5) !}{(n+5-n-1) !} =\frac{11(n-1)}{2} \times \frac{(n+3) !}{(n+3-n) !} \)
\(\frac{(n+5)(n+4)(n+3) !}{4 !} =\frac{11(n-1) \times(n+3) !}{2 \times 3 !} \)
\(\frac{(n+5)(n+4)}{4 \times 3 !} =\frac{11(n-1)}{2 \times 3 !} \)
\(n^{2}+9 n+20 =22 n-22 \)
\(n^{2}-13 n+42 =0 \)
\((n-6)(n-7) =0 \)
\(n=6 \text { or } 7 \)

3 boys can be selected from 5 boys in ⁵C₃ ways 2 girls can be selected from 4 girls in ⁴C₂ ways and 1 transgender can be selected from 2 transgenders in ²C₁ ways.
=\(\frac { 5! }{ 2!3! } \times \frac { 4! }{ 2!2! } \times 2=\frac { 5\times 4\times 3! }{ 2\times 3! } \times \frac { 4\times 3\times 2! }{ 2!2! } \times 2\)
= 10 \(\times\) 6 \(\times\) 2 = 120

5 cars can be chosen out of 100 cars in ¹⁰⁰C₅ ways
=​​​​
= 451725120

A chord is obtained by joining any two points on a circle
Number of chords drawn though 20 points is same as the number of ways of selecting 2 points out of 20 points.
This can be done in 20C₂ ways.
Hence, total number of chords is 20C₂ ​​​​​
=\(\frac { 20! }{ 2!18! } =\frac { 20\times 19\times 18! }{ 2\times 18! } =\frac { 20\times 19 }{ 2 } \)
=\(10\times 19\) = 190.

The total number of handshakes is same as the number of ways of selecting 2 persons among 15 persons.
This can be done in 15C₂ ways
∴ Number of handshakes = \(15{ C }_{ 2 }=\frac { 15! }{ 2!13! } =\frac { 15\times 14\times 13! }{ 2!13! } =\frac { 15\times 14 }{ 2 } =15\times 7\)
= 105.

Here 7 players must be selected from 14 players. This can be done in 14C₇ ways.
Hence, number of different teams of players
= \(14{ C }_{ 7 }=\frac { 14! }{ 7!7! } \)
= \(\frac { 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7! }{ 7!7\times 6\times 5\times 4\times 3\times 2\times 1 } \)
= \(13\times 11\times 2\times 3\times 4\)
= 3432.