Here ^{n - 1}P₃ : ⁿP₄ = 1 : 9
\(\therefore {(n-1)!\over(n-4)!}:{n!\over(n-4)!}=1:9\)
\(\Rightarrow {(n-1)!\over(n-4)!}\times{(n-4)!\over n!}={1\over9} \Rightarrow {(n-1)!\over n!}={1\over9} \Rightarrow {(n-1)!\over n(n-1)!}={1\over9}\)
\(\Rightarrow {1\over n}={1\over9}\Rightarrow n=9\)

(a) 4 letters are used at a time = ⁶p₄ = \(6!\over 2!\) = 360
(b) All letters are used at a time = ⁶p₆ = 6! = 720
(c) All letters are used but first letter is vowel = 2 \(\times\) 5! = 2 \(\times\) 120 = 240
Hence, the required matching is (a)\(\leftrightarrow\)(iii), (b) \(\leftrightarrow\) (i), (c) \(\leftrightarrow\) (ii)

There are 10 questions in Part-A, out of which 8 questions can be chosen in 10C₈ ways.
Similarly from Part-B, contains using 10 questions, 5 questions can be chosen in 10C₅ ways.
Hence, the total number of ways of selecting 8 questions from Part-A and 5 from Part-B.
= 10C₈ \(\times\)10C₅ = \(\frac { 10! }{ 8!2! } \times \frac { 10! }{ 5!\times 5! } \)

= 11340.

The number of arrangements in which the best and the worst papers never come together can be obtained by subtracting from the total number of arrangements, the number of arrangements in which the best and worst come together.
Number of arrangements of 9 papers = 9P₉ = 9!
Considering the best and the worst paper as one paper, we have 8 papers which can be arranged in 8P₈ = 8! ways.
But the best and worst papers can be put together in 2! ways.
So, the number of permutations in which the best and worst papers can be put together = 2! \(\times\) 8! ways.
Hence, the number of ways in which the best and the worst papers never come together.
= 9! - 2! 8!
=9 \(\times\) 8! - 2! 8! =8! (9 - 2)   [\(\because\) 9! = 9\(\times\)8!]
= 8! \(\times\) 7 = 8 \(\times\) 7 \(\times\) 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 \(\times\) 7
= 282240

Given (n+2)! = 60(n-1)!

\(\Rightarrow \) (n+2) (n+1)(n) = 60
\(\Rightarrow \) (n+2)(n+1)(n) = 5 \(\times\)4 \(\times\)3
Equating the terms both sides we get, n = 0