There are 2 roads connecting Chennai to Trichy. Suppose these are R₁ and R₂. Further, there are 3 roads connecting Trichy to Tirunelveli. Let us name them as S₁, S₂ and S₃. Suppose a person chooses R₁ to travel from Chennai to Trichy and may further choose any of the 3 roads S₁, S₂ or S₃ to travel from Trichy to Tirunelveli. Thus the possible road choices are (R₁, S₁), (R₁, S₂), (R₁, S₃). Similarly, if the person chooses R₂ to travel from Chennai to Trichy, the choices would be (R₂, S₁), (R₂, S₂), (R₂, S₃).

Thus there are 2 \(\times\) 3 = 6 ways of travelling from Chennai to Tirunelveli.

Given factors are two a's, 3b's and 4c's
Total number of exponents = 9.
Hence, number of ways the product can be expressed without exponents
= \(\frac { 9! }{ 2!3!4! } =\frac { 9\times 8\times 7\times 6\times 5\times 4\times ! }{ 2\times 3\times 2\times 4! } \)= 1260

There are 6 letters in the word 'SIMPLE'.
So, total number of words is equal to the number of arrangements of these letters, taken all at a time. Sum order of such arrangements is 6 P₆ = 6! = 720

4 coats can be given to 3 men in 4 p₃ ways.5 waist coats can be given to 3 men in 5P₃ ways 6 caps can be given to 3 men in 6 P₃ ways.
∴ Total number of ways of wearing them 
= 4P3 \(\times\) 5P₃ \(\times\) 6 P₃ 
= \(\frac { 4! }{ 1! } \times \frac { 5! }{ 2! } \times \frac { 6! }{ 3! } \)
= \(4\times 3\times 2\times \frac { 5\times 4\times 3\times 2! }{ 2! } X\frac { 6\times 5\times 4\times 3! }{ 3! } \)
= 24 \(\times\) 60 \(\times\) 120 
= 1,72,800

Gold medal can be awarded to anyone of the 8 candidates in 8 ways.
Silver medal can be awarded to anyone of the remaining 7 candidates in 7 ways.
Bronze medal can be awarded to anyone of the remaining 6 candidates in 6 ways.
∴ Total numbers of ways of awarding the prize
= 8 \(\times\) 7 \(\times\) 6 = 336

Given ^(n-1)P₃ :ⁿ P₄ = 1 : 10 
⇒ \(\frac { (n-1){ P }_{ 3 } }{ n{ P }_{ 4 } } =\frac { 1 }{ 10 } \)
⇒ 10.(n-1)P₃ = 1.nP₄  \(\left[ \because npr\quad =\frac { n! }{ (n-r)! } \right] \)
⇒ \(10\times \frac { (n-1)! }{ (n-1-3)! } =\frac { n! }{ (n-4)! } \)
⇒   \(\frac { 10\times (n-1)! }{ (n-4)! } =\) \(\frac { 10\times (n-1)! }{ (n-4)! }\)⇒\(\frac { n(n-1)! }{ (n-4)! } \)
⇒10 = n
∴ n = 10.

Let p(n) = 1 + 3 + 5 + ... + (2n- 1). Therefore P(1) = 1 = 1² is true.
We assume that P(k) = 1 + 3 + 5 +: .. (2k-1) is true for n = k.That is P(k) = k²
We need to prove P(k + 1) = (k+ 1)²
P(k+1) = 1+3+5+...(2(k+1)-1)
\(=\underbrace{1+3+5+7+.....+(2k+1)}+2k+1\)
= P(k) + 2k+1
= k²+2k+1 = (k+1)²
This implies, P(k + 1) is true. Hence, by the principle of a mathematical induction, P(n) is true for all natural numbers.

The given strings contains 6 letters (V, O, W, E, L, S).
(i) Since all strings must begin with E, we have the remaining 5 letters which can be arranged in ⁵P₅ = 5! ways.
Therefore the total number of strings with E as the starting letter is 5! =120.

(ii) Since all strings must begin with E, and end with W, we need to fix E and W. The remaining 4 letters can be arranged in ⁴P₄ = 4! Ways.

Therefore the total number of strings with E as the starting letter and W as the final letter is 4! = 24.

There are three conditions as follows:
1. It is 4-digit number and hence its 1000^th place cannot be 0.
2. It is an even number and hence its unit place can be either 0, 2 or 4.
Two cases arise in this situation. Either 0 in the unit place or not.
Case 1: When the unit place is filled by 0, then the 1000^th place can be filled in 4 ways, 100^th place can be filled in 3 ways and 10^th place in 2 ways. Therefore, number of 4-digit numbers having 0 at unit place is \(4 \times 3 \times 2 \times 1=24\)

Case 2: When the unit place is filled with non-zero numbers, that is 2 or 4, the number of ways is 2, the number of ways of filling the 1000^th place is in 3 ways (excluding '0'), 100^th place in 3 ways and 10^th place in 2 ways. Therefore, number of 4-digit numbers without 0 at unit place is \(3 \times 3 \times 2 \times 2=36\)
Hence, by the rule of sum, the required number of 4 digit even numbers is 24 + 36 = 60.

The following figure illustrates the electricity distribution network.

There are 110 smaller capacity transformer attached to a larger capacity transformer. As each smaller capacity transformer can be linked with only 100 consumers, we have for the 109 transformers, there will be 109 \(\times\)100 = 10900 links. For the 110^th transformer, there are only 29 consumers linked. Hence, the total number of consumer linked to the 238^th larger capacity transformer is 10900 + 29 = 10929.

Lexicographic order of the word is F, N, N, D, Y
Number of words starting with F = \(\frac { 4! }{ 2! } \)
\(\frac{0 \times 4 !}{2 !}+\frac{2 \times 3 !}{2 !}+\frac{0 \times 2 !}{2 !}+0 \times 1 !+0 \times 0 !+1\)
\(=0+\frac{2 \times 3 \times 2 \times 1}{2}+0+0+0+1\)
= 6 + 1 = 7
Rank of FUNNY = 7

(i) When coin is tossed 1 time, Possibilities of heads and tails = 1¹ = 1
When the coin is tossed 2 times, Possibilities of head and tails is 24.
In this way we get,
When a coin tossed at 8 times, possibilities of head and tail of different sequences is 2⁸
(ii) Since there are 6 heads of one kind and 4 tails of other kind, required number of sequences = \(\frac { { 2 }^{ 8 } }{ 6!2! } \)

Considering all S as one letter there are 5 letters containing 2 C's, one U, and one E which can be arranged in
\(\frac { 5! }{ 2!1!1! } =\frac { 5\times 4\times 3\times 2 }{ 2 } \)
= 60 Ways

There are 11 letters in the given word of which 4 are S's,4 are I's and 2 are P's and 1 M
Hence total number of distinct words
= \(\frac { 11! }{ 4!4!2!1! } =\frac { 11! }{ 4!4!2! } \)
=\(\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 1\times 4!\times 2 } \) = 34650

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) Since each question can be answered in 4 ways, the maximum number of different answers
= \(4\times 4\times 4\times 4\times 4={ 4 }^{ 5 }\)
(ii) When each question has more than one correct answer the maximum number of different answers \(=(5 \times 3)^{5}=15^{5}\)

(i) Since each question can be answered in 4 ways, the total number of ways of answering 10 questions is 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 = 4¹⁰
(ii) Since first four question have three choices, the number of ways of answering first four questions is \(3\times 3\times 3\times 3={ 3 }^{ 4 }\)
Remaining 6 questions have 5 choices each
∴ Number of ways of answering the remaining 6 question =\(5\times 5\times 5\times 5\times 5\times 5={ 5 }^{ 6 }\)
∴ Total number of ways of answering the questions = \({ 3 }^{ 4 }\times { 5 }^{ 6 }\)
(iii) Since first four question have three choices, the number of ways of answering first four questions is \(3\times 3\times 3\times 3={ 3 }^{ 4 }\)
Remaining 6 questions have 5 choices each
∴ Number of ways of answering the remaining 6 question =\(5\times 5\times 5\times 5\times 5\times 5={ 5 }^{ 6 }\)
∴ Total number of ways of answering the questions =  \({ 3 }^{ 4 }\times { 5 }^{ 6 }\)

Given ¹⁰P_r-1 = 2 \(\times\) ⁶P_r
⇒ \(\frac { 10! }{ (10-r+1)! } =2\times \frac { 6! }{ (6-r)! } \)   \(\left[ \because n{ P }_{ r }=\frac { n! }{ (n-r)! } \right] \)
⇒ \(\frac { 10\times 9\times 8\times 7\times 6! }{ (11-r)! } =\frac { 10\times 9\times 8\times 7\times 6! }{ (11-r)! } \)
⇒ \(\frac { 10\times 9\times 8\times 7 }{ (11-r)(10-r)(8-r)(7-r)(6-r) } =\frac { 2 }{ (6-r)! } \)
⇒ \(\frac { 10\times 9\times 8\times 7 }{ (11-r)(10-r)(8-r)(7-r) } =2\)
\( \Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r)=5 \times 9 \times 8 \times 7 \)
\( \Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r)=7 \times 6 \times 5 \times 4 \times 3 \)
\( \Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r)=(11-4)(10-4)(9-4)(8-7)(7-4) \)
⇒ r = 4

Let, P(n): = (r(cos ፀ + i sin ፀ)n = rⁿ (cos nፀ+i sin nፀ)
Substituting the value of n = 1, in the statement we get,
P(1) =(r(cos ፀ + i sin ፀ))¹ = r(cos ፀ+i sin ፀ)
Hence, P(1) is true.
Let us assume that the statement is true for n = k. Then
(r(cos ፀ + i sin ፀ))^k = r^k (cos kፀ + i sin kፀ)
We need to show that P(k + 1) is true. Consider,
P(k+1) = ((r(cos ፀ + i sin ፀ))^k+1
=(r(cos ፀ + i sin ፀ))^k+1
=(r(cos ፀ + i sin ፀ))^k \(\times\) r(cosፀ+ isinፀ)
=r^k(coskፀ+i sin kፀ) \(\times\) r(cosፀ+ i sinፀ)
=r^k+1 x (cos kፀ cos ፀ + i² sin kፀ sin) + i(sin kፀ cos ፀ + cos kፀ sin ፀ)
=r^k+1 x (cos(k+1)ፀ + i sin(k+1)ፀ).
This implies that P(k + 1) is true. The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction, for any natural number n,
(r(cos ፀ + i sin ፀ))ⁿ = rⁿ(cos(nፀ) + i sin(nፀ))

There are 11 letters in the word, with respect to number of repetitions of letters there are 4 distinct letters (R, S, T, N), 2 sets of two alike letters (PP, II), 1 set of three alike letters (OOO). The following table will illustrate the combination of these sets and the number of words.

Hence, the total number of strings are 2520 + 20 + 300 + 450 + 3600 = 6890.

(i) There are no restrictions: Totally there are 8 questions in both Part A and Part B. The total number of ways of attempting 5 questions from 8 questions is ⁸C₅ = ⁸C₃ = 56.
(ii) At least two questions from Part A needs to be answered: Accordingly, various choices are tabulated as follows.

Therefore, the required number of ways of answering is
⁴C₂ \(\times\) ⁴C₃ + ⁴C₃ \(\times\) ⁴C₂+ ⁴C₄ \(\times\) ⁴C₁ = 24 + 24 + 4 = 52.

⁽ⁿ⁺²⁾C₇ : ^(n-1)P₄ = 13 : 24
\(\frac { { (n+2) }_{ C_{ 7 } } }{ { (n+2) }_{ P_{ 4 } } } =\frac { 13 }{ 24 } \)
\(\frac { (n+2)! }{ (n-5)!7! } \times \frac { (n-5)! }{ (n-1)! } =\frac { 13 }{ 24 } \)
\(\frac { (n+2)(n+1)n(n-1)! }{ (n-1)!.7! } =\frac { 13 }{ 24 } \)
\((n+2)(n+1)(n)=\frac{13}{24} \times 7 !=\frac{13}{24} \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
\((n+2)(n+1)(n)=13 \times 14 \times 15\)
n + 2 = 15 \(\Longrightarrow\) n = 13.

The lexicographic order of the letters of given word is E, E, I, I, J, T. In the lexicographic order, the strings which begin with E come first. If we fill the first place with E, remaining 5 letters (E, I, I, J, T) can be arranged in \(\frac { 5! }{ 2! } \) ways. On proceeding like this we get,
E - - - - -  = \(\frac { 5! }{ 2! } \) = 60 ways
IE - - - = 4! = 24 ways
IIE - - - = 3! = 6 ways
IIJ - - - = \(\frac { 3! }{ 2! } \) = 3 ways
IITE - - = 2! = 2 ways
IITJEE = 1 way
The rank of the word IITJEE is 60 + 24 + 6 + 3 + 2 + 1 = 96.

(i) As there 8 seats to be occupied out of which one seat is for the one who drives. Since there are no restrictions any one can drive the van. Hence the number of ways of occupying the driver seat is ⁷P₁ = 7 ways. The number of ways of occupying the remaining 7 seats by the remaining 6 people is ⁷P₆ = 5040.
Hence the total number of ways the family can be seated in the car is 7 \(\times\) 5040 = 35280.
(ii) As the driver seat can be occupied by only F or M, there are only two ways it can be occupied. Hence the total number of ways the family can be seated in the car is 2 \(\times\) 5040 = 10080.
(iii) As there are only 5 window seats available for D₁ & D₂ to occupy the number of ways of seated near the windows by the two family members is ⁵P₂ = 20. As the driver seat is occupied by F, the remaining 4 people can be seated in the available 5 seats in ⁵P₄ = 120. Hence the total number of ways the family can be seated in the car is 20 \(\times\) 1 \(\times\) 120 = 2400.

(i) There are 8 letters in the word "EQUATION" which includes 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). Considering 5 vowels as one letter, we have 4 letters which can be arranged in ⁴P₄= 4! ways. But corresponding each of these arrangements, the vowels E, U, A, I, O can be put in ⁵P₅ = 5! ways.
Hence, by the rule of product required number of words is 4! \(\times\) 5! = 24 \(\times\) 120 = 2880.
(ii) The total number of strings formed by using all the eight letters of the word "EQUATION" is ⁸P₈ = 8! = 40320.
So, the total number of strings in which vowels are never together is the same as the difference between the total number of strings and the number of strings in which vowels are together is 40320 - 2880 = 37440.

(i) The solution for this is the same as the number of permutations taking four-digits out of 5 digits is ⁵P₄ = 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 = 120.

(ii) For even number last digits must be 2 or 4 which is filled in ²P₁ ways and remaining 3 places filled from remaining 4 digits in ⁴P₃ ways. Therefore the required number of ways is ²P₁ \(\times\) ⁴P₃ = 2 \(\times\) 24 = 48.

(iii) Since the number divisible by 4, then last two digit must be divisible by 4. The Last two digits become 12, 24, 32, 52 (4 ways). The remaining first two places filled from remaining 3 digits in ³p₂ ways. The required number of numbers which are divisible by 4 is ⁴P₁ \(\times\) ³P₂ = 4 \(\times\) 6 = 24.

n(A) = Number of numbers divisible by 2 = 500
n(B) = Number of numbers divisible by 5 = 200
n(A⋂B) = Number of numbers divisible by 10 = 100
n(AUB) = n(A) + n(B) - n(A⋂B)
Number of numbers divisible by 2 or 5
= 500 + 200 - 100 = 600
∴ Number of numbers divisible neither by 2 nor by 5
= 1000 - 600 = 400

\(\frac { (2n)! }{ n! } =\frac { 1.2.3.4...(2n-2).(2n-1)2n }{ n! } \)
= \(\frac { (1.3.5...(2n-1))(2.4.6...(2n-2).2n) }{ n! } \) (Grouping the odd and even numbers separately)
= \(\frac { (1.3.5...(2n-1))\times 2^{ n }\times (1.2.3...(n-1).n) }{ n! } \) (taking out the 2's)
= \(\frac { (1.3.5...(2n-1)\times 2^{ n }\times n! }{ n! } \)
= 2ⁿ(1.3.5...(2n - 1))

(i) There are as many strings as filling the 4 vacant places by the 4 letters, keeping in mind that repetition is not allowed. The first place can be filled in 4 different ways by any one of the letters B, I, R, D. Following which, the second place can be filled in by any one of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways, following which fourth place can be filled in 1 way. Thus the number of ways in which the 4 places can be filled, by the rule of product is 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24. Hence, the required number of strings is 24.
(ii) There are 5 different letters with which 5 places are to be filled. The first place can be filled in 5 ways as any one of the five letters P, R, I, M, E can be placed there. Having filled the first place with any of the 5 letters, 4 letters are left to be placed in the second place, three letters are left for the third place and 2 letters are left to be put in the fourth place. The remaining 1 letter has to be placed in the fifth place. Hence, the total number of ways filling up five places is 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 120.

Since 0 cannot be in the thousand's place
Let us find the sum of all these 500 numbers.
By filling 0 is the unit place, the remaining 3 places can be filled with remaining 4 digits is \(4\times 4\times 4=64\)  in unit place.
∴ Sum of all the unit digits
= \((64\times 0)+(64\times 2)+(64\times 5)+(64\times 7)+(64\times 8)\)
= 64(0 + 2 + 5 + 7 + 8) = 64 (22) = 1408
Similarly sum of the tens digits =\(14408\times 10\) = 144080
Sum of all the hundred's digits =  \(14408\times 100\) = 1440800
By filling 2 in the thousand's place, remaining 3 places can be filled with remaining 3 digits in \(3\times 3\times 3\) 27 ways [since 0 cannot be in thousands place]
∴ Sum of the digits in the thousands place
= \(27(2+5+7+8)=27(22)\times 1000\) = 594000
Hence sum of all the 4-digit numbers formed by using the digits 0, 2,5, 7, 8 is 1408 + 14080 + 140800 + 594000
= 1408 (1 + 10 + 100) + 594000
= \(1408\times 11+594000\)
= 750288

The number of 4-digit numbers that can be formed using the 5 digits is ⁵P₄ = 120
Let us find the sum of the digits in the unit place.
 \(=5 \times 4 \times 3 \times 2=20 \times 6\)
= 120 numbers
Each of the five given numbers will be repeated 24 times. Hence, sum of digits appearing in any place
= 24 (1 + 2 + 3 + 4 + 5)
24 \(\times\) 15 = 360
Sum of all 4 digits = 360 (1000 +100 + 10 + 1)
360 \(\times\) 1111 = 399960

In the word THING, there are 5 letters
The lexicographic order of the word is G, H, I,N, T
Number of words starting with G = 4! = 24
Number of words starting with H =  4! = 24
Number of words starting with I 4! = 24
Number of words starting with NG 3! = 6
Number of words starting with NGH 2! = 2
Number of words starting with NGHI = 1!
Number of words starting with NGHIT = 1!
85^th string NGHIT 

(i) The lexicographic order of the letters of the given word is A, D, E, G, N, R.
The word GARDEN has 6 letters in which no letters are repeating.
∴ Rank \( = (3 \times 5 !)+(0 \times 4 !)+(3 \times 3 !)+ (0 \times 2 !)+(0 \times 1 !)+(0 \times 0 !)+1 = 3 \times 120+0+18+0+0+0+1 \)
 = 360 + 19 = 379
(ii) The lexicographic order of the letters of the given word is A, D, E, G, N, R.
The word GARDEN has 6 letters in 'which no letters are repeating
 Rank = \((1 \times 5 !)+0+(2 \times 3 !)+(1 \times 2 !)\)
 ∴  Rank of DANGER = 135.

(i) Given numbers are 1, 1, 2, 3, 3, 4
Here 1 occur twice
3 occur twice
∴  Number of distinct 6-digit numbers

(ii) Unit place can be filled in 2 ways by 2 or 4.
Remaining 5 digits can be filled in \(\frac { 5! }{ 2!2! } \)
∴  Number of even 6 digit numbers
 = \(\frac { 5! }{ 2!2! } \times 2\)
= \(\frac { 5\times 4\times 3\times 2 }{ 2! } \) = 60
(iii) Unit place can be filled in one ways by 4.
Remaining 5 digits can be filled in \(\frac { 5! }{ 2!2! } \) ways
∴ Number of even 6 digit numbers divisible by 4
= \(\frac { 5! }{ 2!2! } \times 1\)
= \(\frac { 5\times 4\times 3\times 2\times 1 }{ 2\times 2 } \)  = 30

(i) In 'INTERMEDIATE' the vowels are I, E, E, I, E, A and there are 12 letters. 
Here their 3 E's. 2' 1's and 1A
Total number of arrangement
 = \(\frac { 12! }{ 3!2!1! } =\frac { 12\times 11\times 10\times 9\times 8\times 7+6\times 5\times 4\times 3! }{ 3\times 2\times 2 } \)
= \(\frac { 479001600 }{ 2 } \) = 39916800
(ii) These 6 vowels can be arranged among themselves in 5! ways considering these vowels as one unit, and the remaining 6 consonants, required number of arrangement = \(7!\times 6!\)
(iii) The number of letters in the word INTERMEDIATE = 12 out of which 6 are vowels and 6 are consonants. 

(iv) 6 consonants out of which 2 are alike can be place in \(\frac { 6! }{ 2! } \) ways and 6 vowels out of which 3E's are alike and 21's are alike can be arranged in 7 places in \({ 7P }_{ 6 }\times \frac { 1 }{ 3! } \times \frac { 1 }{ 2! } \) ways 
∴ Total number of words = \(\frac { 6! }{ 2! } \times { 7P }_{ 6 }\times \frac { 1 }{ 3! } \times \frac { 1 }{ 2! } \)
= \(\frac { 6\times 5\times 4\times 3! }{ 2 } \times \frac { 7! }{ 1! } \times \frac { 1 }{ 3! } \times \frac { 1 }{ 2 } \)
= \(30\times 7\times 6\times 5\times 4\times 3\times 2\times 1\)
= 151200

Four subjects can be arranged on the shelf in 4! ways.
The books on mathematics can be arranged in 4! ways, physics on 3! ways, chemistry on 2! ways and Biology on 1! ways.
Hence, total number of ways of arranging the books
= 4! 4! 3! 2! 1!
= \((4\times 3\times 2\times 1)(4\times 3\times 2\times 1)(3\times 2)(2\times 1)\)
= (24)(24)(6)(2)
= 6912

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