The number of combination = \({ 32C }_{ 4 }\) 
\( =\cfrac { 32! }{ 4!(32-4)! } \)
\(=\cfrac { 32! }{ 4!\left( 28 \right) ! } \)

When the students are arranged along a circle, then the number of permutation is (8 – 1)! = 7!

8P₃ = 8 \(\times\) 7 \(\times\) 6 = 336.

We have P(n): 2³ⁿ-1 is a multiple of 7
 ∴  P (5) : 2¹⁵ -1 = 32767
= 7 (4681) which is multiple of 7
\(\therefore\) P (5) is true

An octagon has 8 angular points 
∴ Number of lines = 8 C₂ =\(\frac { 8\times 7 }{ 1\times 2 } =27\)
Number of sides = 8
∴ Number of diagonals 28 - 8 = 20

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

Number of vowels are A, E, I, 0, U = 5
∴ Taking 2 vowels from 5 vowels = 5P₂
= 5 x 4 = 20

LHS 10P₃ = 10 x 9 x 8 = 720
RHS 9P₃ + 3. 9P₂ = 9 x 8 x 7 + 3 x 9 x 8
= 9 x 8 (7 + 3) = 72 (10) = 720
LHS= RHS  Hence proved.

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

Four dice are rolled, total number of possible outcomes = 6⁴
Number of possible outcomes in which 2 does not appear on any dice = 5⁴
∴ Required number of possible outcomes = Total number of possible outcomes - number of possible outcomes in which 2 does not appear on any dice.
= 6⁴ - 5⁴ = 1296 - 625
= 671

\(\frac { 7! }{ 6! } =\frac { 7\times 6! }{ 6! } =7\)

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

\(\left( n+2 \right) { C }_{ n }=45\)
\(\left( n+2 \right) { C }_{ n+2-n }=45\)
\( \left( n+2 \right) { C }_{ 2 }=45\)
\(\cfrac { \left( n+2 \right) \left( n+1 \right) }{ 2 } =45\)
\({ n }^{ 2 }+3n-88=0\)
\(\left( n+11 \right) \left( n-8 \right) =0\)
\( n=-11,8\)
\(n=-11\) is not possible
\(\therefore n=8\)

If \(n{ C }_{ x }=n{ C }_{ y }\) then x + y = n
Here \(n{ C }_{ 4 }=n{ C }_{ 6 }\) 
\(\therefore n=4+6=10\)
\({ 12C }_{ n }=12{ C }_{ 10 }\)
\(={ 12C }_{ 2 }\)
\(=\cfrac { 12\times 11 }{ 1\times 2 } =66\)

Here, n = 9 and r = 3 `
\(\therefore\) Number of 3 digit numbers = n^r
= 9³ = 729 numbers