(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.

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 

Here total number of digits = 6
The unit place can be filled with any one of the digits 2, 4, 6.
So number of permutation = ⁵P₁ = \({3!\over 2!}=3\)
Now the tens and hundreds place can be filled by remaining 5 digits.
So number of permutation = ⁵P₁= \({5!\over3!}={5\times 4\times 3!\over 3!}=20\)
Hence total number of permutations = 3 \(\times\) 20 = 60

The alphabetical order of RACHIT is A, C, H, I, Rand T
Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with I = 5!
and Number of words beginning with R (i.e) RACHIT = 1
\(\therefore\)The rank of the word 'RACHIT' in the dictionary
= 5! + 5! + 5! + 5! + 1 = 4 \(\times\) 5! + 1
= 4 \(\times\) 5 \(\times\)4\(\times\)3 \(\times\)2\(\times\)1 + 1 = 4 \(\times\) 120 + 1 = 480 + 1 = 481

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.

Given p(n) : "n²-n + 41 is prime".
Now p(1) = 1²-1+41 = 41 which is a prime number.
\(\Rightarrow\) p(1) is true.
p(2) = 2² - 2 + 41 = 4 - 2 + 41 = 43 which is a prime number.
\(\Rightarrow\) p(2) is true.
Now, p(3) = 3²- 3 + 41
= 9 - 3 + 41 = 47 which is also a prime number.
\(\Rightarrow\) p(3) is true.
Consider p(41) = 41²- 41 + 41 = 41² =1681 which is not a prime since it is divisible by 41.

p(n): "n² + n is even"
Given p(r) is true
\(\Rightarrow\) r²+ r is even
\(\Rightarrow\) r²+ r = 2k where k is a constant
To prove that p(r+1) is true
To prove that (r+1)² + (r+1) is even
Consider (r+1)²+r+1
=r²+2r+1+r+1
= (r²+ r) + 2r + 2
= 2k + 2r + 2
= 2(k+r+1) which is even always
\(\therefore\) p(r+1) is true.

Since 4 particular guest want to sit on side A and 3 on the other side B, so we are left with 11 guests out of which we choose 5 for side in 11C₅ ways and 6 for side B in 6C₆ ways.
\(\therefore\) Number of selections for the two sides is 11C₅\(\times\)6C₆.
Now, 9 persons on each side of the table can be arranged among themselves in 9! ways.
Hence, the total number of arrangement.
= 11C₅ \(\times\) 6C₆\(\times\) 9! \(\times\) 9!