6 gentlemen can be arranged in a circle in (6 - 1)! = 5! = 120 ways
The ladies can occupy the X marked seats
Number of ways of arranging ladies = 6 P₄ = 6 x 5 x 4 x 3 = 360
∴  Total number of arrangements = 120 x 360 = 43200.

In "AGAIN" the letters in ascending order are A, A, G, I, N
∴  The first word is AAGIN
Number of words beginning with A A
= No of ways of arranging G, 1, N = 3! = 6
The next word begin with AG and it is AGAIN
∴  No. of words before AGAIN = 6.
∴  Rank of word AGAIN = 7

Any number between 100 and 1000 is a 3 - digit number

1. Hundred place can be filled in 6 ways using the numbers 1,2,3,4,5,6 (excluding 0)
2. Tens place can be filled in 6 ways (including 0)
3. Ones place can be filled in 5 ways.
\(\therefore\)  By fundamental principle of counting, total number of 3 digit numbers = 6 x 6 x 5 = 180.

\({{(2x+1)!}\over{(x+2)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\)  \({{(2x+1)(2x)(2x-1)!}\over{(x+2)(x+1)(x-1)!}}.{{(x-1)!}\over{(2x-1)!}}={{3}\over{5}}\)
\(\Rightarrow\)  \({{(2x+1)(2)}\over{(x+2)(x+1)}}={{3}\over{5}}\)
\(\Rightarrow\)   10 (2x+1) = 3 (x+2) (x+1)
\(\Rightarrow\)   20x+10=3 (x²+3x+2)
\(\Rightarrow\)   20x+10+3x²-9x-6=0
\(\Rightarrow\)   -3x²+11x+4 = 0
\(\Rightarrow\)   -3x²-11x-4=0
\(\Rightarrow\)   (x-4)(3x+1)=0
\(\Rightarrow\)   x = 4 or x \(={{-1}\over{3}}\)
Since \(x={{-1}\over{3}}\) is not possible, x = 4.

\({{x+4}\over{(x^2-4)(x+1)}}={{x+4}\over{(x+2)(x-2)(x+1)}}={{A}\over{x+2}}+{{B}\over{x-2}}+{{C}\over{x+1}}\)
\(\Rightarrow\)  \({{x+4}\over{(x^2-4)(x+1)}}=\frac { { { A(x-2)(x+1)B+(x+2)(x+1)+C(x+2)(x-2) } } }{ (x+2)(x-2)(x+1) } xx\)
\(\Rightarrow\)  x + 4 = A (x - 2) (x + 1) + B (x + 2) (x + 1) C (x +2)(x - 2)   ...(1)
Putting x = 2 in (1) we get,
6 = B (4) (3)  \(\Rightarrow\boxed{B={{1}\over{2}}}\)
Putting x = -1 in (1) we get,
3 = C (1)(-3)  \(\Rightarrow\quad \boxed{C=-1}\)
Putting x = 0 in (1) we get,
4 = - 2A + 2B - 4C
\(\Rightarrow\)   \(4=-2A+2\left( {{1}\over{2}} \right)-4(-1)\)     \(\left[ \because B={{1}\over{2}},C=-1 \right]\)
\(\Rightarrow\)   \(4=-2A+1+4\ \ \Rightarrow\ 4=-2A+5\ \Rightarrow -1\)   =-2A
\(\Rightarrow\)   \(\boxed{A={{1}\over{2}}}\)
\(\therefore\) \({{x+4}\over{(x^2-4)(x+1)}}=​​​​​​{{{{1}\over{2}}}\over{x+2}}​​+{{{{1}\over{2}}}\over{x-2}}-{{1}\over{}x+1}={{1}\over{2(x+2)}}+{{1}\over{2(2x-2)}}-{{1}\over{x+1}}\)