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

There are 13 letters in the given word of which 4 are S's, 2 are I's , 3 A's, 2N's 1 - 0 and 1 - T.
No. of arrangements \(=\frac{13 !}{4 ! 3 ! 2 ! 2 !}\)

\(\left(3 x+\frac{x^2}{2}\right)^8\)
n = 8
Middle term is \(t_{\frac{n}{2}+1}=t_{\frac{8}{2}+1}=t_5\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
r = 4
\(t_5=8 c_4(3 x)^4\left(\frac{x^2}{2}\right)^4\)
\(=\frac{8 C_4(81)}{16} x^{12}\)

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

(999)⁵ = (1000 - 1)⁵
\(=5 C_0(1000)^5-5 C_1(1000)^4+5 C_2(1000)^3 -5 C_3(1000)^2+5 C_4(1000)-5 C_5\)
= 1000,000,000,000,000 - 5 (1,000,000,000,000) + 10 (1,000,000,000) -10 (1000,000) + 5 (1000)-1
= 1000000000000000 - 5000000000000 + 10000000000 - 10000000+ 5000 1
= 99 500 999 000 4999

(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

15C_3r = 15C_{r + 3}
Then by the property,
\(n{ C }_{ x }=n{ C }_{ y }\Longrightarrow x+y=n\) we have 
\(3r+r+3=15\)
\(\Rightarrow r=3\)

nCx = nCy \(\Rightarrow\)x= y or x + y = n
\(\Rightarrow\)  \(\therefore\) 25 _Cn+5 = 25 C_2n-1
\(\Rightarrow\)   n + 5 = 2n-1
\(\Rightarrow\)   n =5 (or) n + 5 + 2n - 1 = 25
\(\Rightarrow\)   6 = 2n - n or 3n + 4 = 25
\(\Rightarrow\) n=  6 or 3n = 21\(\Rightarrow\) n=7
∴ n = 6 or n = 7

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

The 5 boys can be seated in a row in 5P₅ = 5! ways the arrangement can be as follows.
GB GB GB GB GB GB
Since no two girls are to set together we may arrange 3 girls in 6 places which can be done in 6P₃ ways
Total no. of seating arrangement
\(=5 ! \times 6 P_3=5 \times 4 \times 3 \times 2 \times 1 \times 6 \times 5 \times 4\)
= 14400

Since sum of all binomial coefficients is 2n
5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅ + (5C₁ + 5C₂ + 5C₃ + 5C₄)
= 2⁵ + (2⁵ - (5C₀ + 5C₅)
= 32 + 32 - (1 + 1)
= 64 - 2 = 2⁶ - 2

210 = 1024 (2.x 2... 10 terms)

\(\text { M.T } =t_{\frac{n}{2}}+1=t_{5+1}=t_6 \quad r=5 \)
\(t_6 =10 C_5\left(x^5\right)\left(\frac{1}{x^5}\right)=10 C_5 \)

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

Here x² + 1 cannot be factorized into linear factors.
Let \(\cfrac { 2x+1 }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } =\cfrac { A }{ x-1 } +\cfrac { Bx+C }{ { x }^{ 2 }+1 } \) ...(1)
Multiplying both sides by \((x-1)\left( { x }^{ 2 }+1 \right) \) 
\(2x+1=A\left( { x }^{ 2 }+1 \right) +\left( Bx+C \right) \left( x-1 \right) \) ..(2)
Put x =1 in (2)
\(2+1=A(1+1)+0\)
\(\therefore A=\cfrac { 3 }{ 2 } \) 
Put x = 0 in (2)
\(0+1=A(0+1)+(0+C)(-1)\)
\(1=A-C\)
\(\therefore C=\cfrac { 1 }{ 2 } \) 
Equating the coefficient of x² on both the sides of (2), we get
\(A+B=0\)
\(B=-A\)
\( \therefore B=\cfrac { -3 }{ 2 } \) 
Substituting the values of A, B and C in (1), we get
\(\cfrac { 2x+1 }{ (x-1)\left( { x }^{ 2 }+1 \right) } =\cfrac { \frac { 3 }{ 2 } }{ x-1 } +\cfrac { \frac { -3 }{ 2 } x+\frac { 1 }{ 2 } }{ { x }^{ 2 }+1 } \)
\(=\cfrac { 3 }{ 2(x-1) } -\cfrac { 3x-1 }{ 2({ x }^{ 2 }+1) } \)

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\(=\frac { m(m-1) }{ 2\times 1 } \times \frac { n(n-1) }{ 2\times 1 } =\frac { mn(m-1)(n-1) }{ 4 } \)

I and II prizes Mathematics can be given in (30 x 29) ways.
I & II prizes in physics also can be given in (30 x 29) ways.
I prize in chemistry can be given in 30 ways.
I prize in English can be given in 30 ways.
Hence, the number of ways to give prizes in all the four subjects = (30 x 29) x (30 x29)x 30 x 30
= 6.8121 x 10⁸

\(\left(x+\frac{2^{-}}{x^2}\right)^{17}\)
\(n=17\)
\(T_{r+1}=n C_r \cdot x^{n-r} a^{r h}\)
\(=17 C_r x^{17-r}\left(\frac{2}{x^2}\right)^{\mathrm{r}}\)
\(=17 C_r 2^r x^{17-r-2 r}=17 C_r 2^r x^{17-3 r}\)
To find co-efficient of \(x^{11}\) equate the power of x to 11
17 - 3r = 11
17 - 11 = 3r
6 = 3r \(\Rightarrow\) r = 2
Co-efficient of \(x^{11}=17 C_2 2^2\)
\(=\frac{17 \times 16}{2 \times 1} \times 2^2=544\)