Number of words \(=7 C_3 \times 4 C_2 \times 5 !\)
\(=\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} \times 120\)
\(=25200\)
Since out of 7 consonants and 4 vowels 3 consonants and 2 vowels can be chosen in \(7 C_3 \times 4 C_2 \text { ways }\)
Thus there are \(7 C_3 \times 4 C_2\) each containing 3 consonants and 2 vowels. Since each group contains 5 letters which can be arranged among themselves in 5! Ways.

\(\frac { 8! }{ 5! } =\frac { 8\times 7\times 6\times 5! }{ 5! } =336\)

LHS = 8C₄+8C₃
\(=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}+\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\)
\(=70+56=126\)
\(\mathrm{RHS}=9 C_4=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)
\(=126\)
Hence verified

a) When identical beads are arranged along a circle, the number of permutations
\(=\frac { (n-1)! }{ 2 } \)
= \(\frac { (8-1)! }{ 2 } =\frac { 7! }{ 2 } \)
b) When boys are arranged along a circle then the number of permutations
\(=(n-1) !=(8-1) !=7 !\)

\({{x-2}\over{(x+2){(x-1)}^{2}}}={{A}\over{x+2}}+{{B}\over{x-1}}+{{C}\over{{(x-1)}^{2}}}\)
\(\frac{x-2}{(x+2)(x-1)^2}=\frac{A(x-1)^2+B(x+2)(x-1)+C(x+2)}{(x+2)(x-1)^2}\)
X - 2 = A(x - 1)²+ B(x +2)(x - 1)+ C(x + 2)     ..(1)
x = -2 in (1) we get,
-2 - 2 =  A (-3)2 \(\Rightarrow\) - 4 = 9 A \(\Rightarrow\) A = \({{-4}\over{9}}\)
x = 1 in (1) we get,
1- 2 = C(1 + 2) \(\Rightarrow\) -1 = 3C \(\Rightarrow\) C = \({{-1}\over{3}}\)
Equate co-efficient of x² on both sides of (1)
0 = A + B
\(B=-A=\frac{4}{9}\)  
\(\therefore\)   \({{x-2}\over{(x+1){(x-1)}^{2}}}-{{-{{4}\over{9}}}\over{x+2}}+{{{{4}\over{9}}}\over{x-1}}+{{-{{1}\over{3}}}\over{{(x-1)}^{2}}}+{{-4}\over{9(x+2)}}+{{4}\over{9(x-1)}}-{{1}\over{3{(x+1)}^{2}}}\)