Here 12 things are to be divided equally among 4 persons, hence the groups are regarded as distinct.
Hence required number of ways = \(\frac { 12! }{ { (3!) }^{ 4 } } \)
\(=\frac { 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 }{ 3\times 2\times 3\times 2\times 3\times 2\times 3\times 2 } \)
=3696000

No. of ways of giving I prize = n
No. of ways of giving II prize = n
.............................................
Similarly no. of ways of giving n^th prize = n
Therefore By fundamental principle of counting, the required number of ways 

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.

\(n P_4=12\left(n P_2\right)\)
\(n(n-1)(n-2)(n-3)=12 n(n-1)\)
\((n-2)(n-3)=12\)
\(n^2-5 n+6-12=0\)
\(n^2-5 n-6=0\)
\((n-6)(n+1)=0\)
\(n=6\)

\((x-2 y)^{13}\)
\(T_{r+1}=n C_r x^{n-r} a^r\)
\(n=13, r=4\)
\(t_{r+1}=13 C_r x^{13-r}(-2 y)^r\)
\(t_5=13 C_4 x^{13-4}(-2 y)^4\)
\(=\frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} x^9(16) y^4\)
\(=11440 x^9 y^4\)

(101)⁴ = (100 + 1)⁴
= 4C₀ (100)⁴ + 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² + 4 C₃ (100)¹ (1)³ + 4 C₄(1)⁴
= 100,000,000 + 4 (1,000,000) + 6 (10000) + 4 (100) + 1
= 10,40,60,401

\(\left(x+\frac{1}{x^2}\right)^6 = 6 C_0 x^3+6 C_1 x^5\left(\frac{1}{x^2}\right)+6 C_2 x^4\left(\frac{1}{x^2}\right)+6 C_3 x^3\left(\frac{1}{x^2}\right)^3 +6 C_4 x^2\left(\frac{1}{x^2}\right)^4+6 C_5 x\left(\frac{1}{x^2}\right)^5+6 C_6\left(\frac{1}{x^2}\right)^6 \)
\(=x^6+6 x^3+15 x^4\left(\frac{1}{x^4}\right)+20 x^3\left(\frac{1}{x^6}\right)+15 x^2\left(\frac{1}{x^8}\right)+6 x\left(\frac{1}{x^{10}}\right)+\frac{1}{x^{12}}\)
\(=x^6+6 x^3+15+\frac{20}{x^3}+\frac{15}{x^6}+\frac{6}{x^9}+\frac{1}{x^{12}}\)

\(\left(x+\frac{1}{y}\right)^7 =7 C_0 x^7+7 C_1 x^6\left(\frac{1}{y}\right)+7 C_2 x^5\left(\frac{1}{y}\right)^2 +7 C_4 x^3\left(\frac{1}{y}\right)^4+7 C_5 x^2\left(\frac{1}{y}\right)^5 +7 C_6 x\left(\frac{1}{y}\right)^6+7 C_7\left(\frac{1}{y}\right)^7\)
\(=x^7+\frac{7 x^6}{y}+\frac{21 x^5}{y^2}+\frac{35 x^4}{y^3} +\frac{35 x^3}{y^4}+\frac{21 x^2}{y^5}+\frac{7 x}{y^6}+\frac{1}{y^7}\)

A parallelogram is formed by choosing two straight lines from the set of m parallel lines and two straight lines from the set of n parallel lines. Two straight lines from the set of m parallel lines can be chosen in mC₂ ways and Two straight lines from the set of n parallel lines can be chosen in nC₂ ways
Hence, the number of parallelograms formed\(\Rightarrow\)mC₂\(\times \)nC₂\(=\frac { m(m-1) }{ 2\times 1 } \times \frac { n(n-1) }{ 2\times 1 } =\frac { mn(m-1)(n-1) }{ 4 } \)

\({ \left( x+\frac { 1 }{ x } \right) }^{ 2n }\)
t_{r+ 1} = nC_r x^n-r a^r
\(\therefore\) t_r+1 = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-2r
To find the term independent of x, equate the power of x as 0
 2n = 2r \(\Rightarrow\) r = n
\(\therefore\) t_r+1 = 2nC_n.x^2n-2n = 2nC_n
\(={{2n!}\over{n!n!}}={{(2n)(2n-1)(2n-2)(2n-3)......4,3,2,1}\over{n!n!}}\)
\(={{[(2n)(2n-2)(2n-4)...4.2][(2n-1)(2n-3)......3.1]}\over{n!n!}}\)
\(={{2^n(n)(n-1)(n-2).2.1.[(2n-1)(2n-3)....3.1]}\over{n!n!}}\) 
\(=\frac{2^n . n !(2 n-1) \ldots 5.3 .1}{n ! n !}\)
\(=\frac{1.3 .5 \ldots(2 n-1) 2^n}{n !}\) 
Hence proved.

\(\frac { { x }^{ 2 }-3 }{ (x+2)({ x }^{ 2 }+1) } =\frac { A }{ x+2 } +\frac { Bx+C }{ { x }^{ 2 }+1 } \)
[since x² + 1 cannot be factorised into linear factors]
⇒ \(\frac { { x }^{ 3 }-3 }{ (x+2)({ x }^{ 2 }+1) } =\frac { A({ x }^{ 2 }+1)+(Bx+C)(x+2) }{ (x+2)({ x }^{ 2 }+1) } \)
⇒ x² - 3 = A(x² + 1) + (Bx + C) (x + 2)   ...(1)
Putting x = -2 in (1) we get
4 - 3 = A(4 + 1) ⇒ 1 ⇒ 5A ⇒ \(A=\frac{1}{5}\)
Equate co-efficient of x² on both sides of (1)
1 = A + B
\(1=\frac{1}{5}+B\)
\(B=1-\frac{1}{5}=\frac{4}{5}\)
Putting x = 0 in (1) we get,
\(-3=A+2 C\)
\(-3=\frac{1}{5}+2 C\)
\(2 C=-3-\frac{1}{5}=-\frac{16}{5}\)
\(C=-\frac{8}{5}\)
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{\frac{4 x}{5}-\frac{8}{5}}{x^2+1}\)
\(=\frac{1}{5(x+2)}+\frac{4(x-2)}{5\left(x^2+1\right)}\).

\(\frac{2 x^2-5 x-7}{(x-2)^3}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}\)
⇒ \(\frac { 2x^{ 2 }-5x-7 }{ (x-2)^{ 3 } } =\frac { A(x-2)^{ 2 }+B(x-2)+C }{ (x-2)^{ 3 } } \)
⇒ 2x² - 5x - 7 = A(x - 2)² + B(x - 2) + C     ..(1)
x = 2 in (1) we get,
2(4) - 5(2) - 7 = C
C = 8 - 10 - 7 = - 9
Equating co-efficient of x² on both sides of (1)
2 = A
If x = 0
\(-7=A(-2)^2-2 B+C\)
\(-7=2(4)-2 B-9\)
\(-7-8+9=-2 B\)
\(-6=-2 B \Rightarrow B \doteq 3\)
\(\frac{2 x^2-5 x-7}{(x-2)^2}=\frac{2}{x-2}+\frac{3}{(x-2)^2}-\frac{9}{(x-2)^3}\)

\(n P_r =720(\mathrm{nCr}) \)
\(\frac{n !}{(n-r) !} =720\left(\frac{n !}{r !(n-r) !}\right) \)
\(r ! =720 \)
\(r ! =6 ! \)
r = 6

No. of ways = 5C₄ = 5C₁ = 5

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

x + y = n
3 + 2 = 5 = n
nC₄ = 5C₄ = 5C₁ = 5

\(\frac { 1 }{ 5! } +\frac { 1 }{ 6! } +\frac { 1 }{ 7! } \)
\(=\frac { 1 }{ 5! } +\frac { 1 }{ 6\times 5! } +\frac { 1 }{ 7\times 6\times 5! } =\frac { 1 }{ 5! } \left( 1+\frac { 1 }{ 6 } +\frac { 1 }{ 42 } \right) =\frac { 1 }{ 5! } \left( \frac { 42+7+1 }{ 42 } \right) \)
\(=\frac { 1 }{ 5! } \frac { \left( 50 \right) }{ 42 } =\frac { 5 }{ 5! } \)

Number of ways for first passenger to occupy seat = 6. ....(1)
Number of ways for the second passenger to occupy seat = 5
Number of ways for the third passenger to occupy seat = 4.
∴By fundamental principle of counting, the total number of ways for all the three to occupy seats
= 6 x 5 x 4 = 120.

\(\frac { 12x-17 }{ (x-2)(x-1) } =\frac { A }{ x+2 } +\frac { B }{ x-1 } \)
\(\Rightarrow \frac { 12x-17 }{ (x-2)(x-1) } =\frac { A(x-1)+B(x-2) }{ (x-2)(x-1) } \)
\(\Rightarrow 12x-17=A(x-1)+B(x-2)\)
Putting x=1 in (1) we get, 
\(12-17= B(1-2) \Rightarrow -5=-B \Rightarrow \boxed { B=5 } \)
Putting x=2 in (1) we get, 
\(24-17=A(2-1) \Rightarrow 7=A(1) \Rightarrow \boxed { A=7 } \)
\(\therefore \frac { 12x-17 }{ (x-2)(x-1) } =\frac { 7 }{ x+2 } +\frac { 5 }{ x-1 } \)