There are 11 letters in the word MISSISSIPPI
In this word M occurs once I occurs 4 times
S occurs 4 times
P occurs twice.
Therefore required number of permutation
\(=\cfrac { 11! }{ 4!4!2! } \) 

A hexagon has 6 vertices and to draw a triangle we need 3 points
Number of triangles \(=6 C_3=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20\)

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 \(=\frac{13 !}{4 ! 3 ! 2 ! 2 !}\)

\(\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! } \)
\(\frac { 1 }{ 6! } +\frac { 1 }{ 7\times 6! } =\frac { x }{ 8\times 7\times 6! } \)
\(\frac{1}{6 !}+\left(1+\frac{1}{7}\right)=\frac{x}{8 \times 7 \times 6 !}\)
\(\frac{8}{7}=\frac{6 ! x}{8 \times 7 \times 6 !}\)
\(x=\frac{8 \times 8 \times 7}{7}=64\)

\(\left(x+\frac{1}{x}\right)^{11}\)
n = 11
There are 2 middle terms namely
\(\frac{t_{n+1}}{2}, \frac{t_{n+3}}{2}=\frac{t_{11+1}}{2}, \frac{t_{11+3}}{2}=t_6, t_7\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(r=5\)
\(t_6=11 C_5 x^6\left(\frac{1}{x}\right)^5=11 C_5 x^6\left(\frac{1}{x^5}\right)\)
\(=11 \mathrm{C}_5 x\)
r = 6
\(t_7=11 C_6 x^5\left(\frac{1}{x}\right)^6=11 C_6 x^5\left(\frac{1}{x^6}\right)\)
\(=\frac{11 C_5}{x}\)

(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

Let P (n) denote the statement 1³ + 2³ + 3³ + ....... + n³ = \(\frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } \)
Put n = 1
LHS = 1³ = 1
\(=\frac { 1^2(2)^2 }{ 4 } \Rightarrow 1\)
LHS = RHS
\(\therefore\) P (1) is true
Let us assume that P(k) is true
p(k) : 1³ + 2³ + ..... + k³ = \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } \)
To prove that P(k+1) IS true
p(k) : 1³ + 2³ + ..... + k³ + (k + 1)³
\(=P(k)+(k+1)^3\)
= \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } +(k+1)^{ 3 }\)
\(=\frac{k^2(k+1)^2+4(k+1)^3}{4}\)
\(=\frac{(k+1)^2\left(k^2+4(k+1)\right)}{4}\)
\(=\frac{(k+1)^2\left(k^2+4 k+4\right)}{4}=\frac{(k+1)^2(k+2)^2}{4}\)
∴ p(k + 1) is true if P(k) is true.
∴ p(n) is true for all \(n\in N\)

\({ \left( x+\frac { 1 }{ x } \right) }^{ 2n }\)
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.

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

nC₀ + nC₁ + nC₂ + nC₃ + ... +nC_n = 2n and nC₀ = 1

Since if n = 1 then (1) (2) (3) (4) = 24 is divisible by = 24

nP₂ = 20
n(n - 1) = 5 x 4
n = 5