There are 11 letters viz MM, AA, TT, H, E, f, C, S. We can choose 4 letters from the following ways:
(i) All the 4 distinct letters: There are g distinct letters viz, M, A, T, H, E, I, C, S out of which 4 canbe chosen in 8C₄ ways. These 4 letters can be arranged in 4! ways
Total number of words = 8C₄ \(\times\) 4!
= 8P₄= 1680
(ii) 2 distinct and 2 alike letters: There are 3 pairs of alike letters viz MM, AA, TT out of which one Pair can be chosen in 3C₁ ways.
Now we have to choose 2 letters out of the remaining 7 letters which can be done in 7C₂ ways.
Each such group has 4 letters of which 2 are alike and remaining 2 distinct and they can be arranged in \(\frac{4 !}{2 !}\) ways
Total number of which 2 letters are alike \(=3 C_{1} \times 7 C_{2} \times \frac{4 !}{2 !}=756\)
(iii) There are 3 pairs of 2 alike letters out of which 2 pats can be chosen in 3C₂ ways. So, there are 3C₂ groups of 4 letters each.
In each groups there ate 4 letters of which 2 are alike of one kind 2 alikeof other kind. These 4 letters can be arcanged in \(\frac{4 !}{{2 !{2 !}}}\) ways.
Hence the total number of words in which 2 letters are alike of one kind and two alike of other kind \(=3 C_{2} \times \frac{4 !}{2 ! 2 !}=18\)
Total number of 4 letters word = 1680 + 756 + 18 = 2454

Let the given statement P(n) be defined as
\(1+2+3+...+n=\cfrac { n\left( n+1 \right) }{ 2 } ,n\varepsilon N\) 
Step 1 : \(n=1\)
\(LHS\ P(1)=1\)
\(RHS\ P(1)=\cfrac { 1\left( 1+1 \right) }{ 2 } =1\) 
LHS = RHS for n = 1 `
\(\therefore\) P(1) is true
Step 2: Let us assume that the statement is true for n = k.
i.e., P(k) is true
\(1+2+3+...+k=\cfrac { k\left( k+1 \right) }{ 2 } \) is true 
Step 3: To prove that P(k + 1) is true
\(p(k+1)=1+2+3+...+k(k+1)\)
\(= p(k)+k+1\)
\(=\cfrac { k(k+1) }{ 2 } +k+1\)
\(=\cfrac { k(k+1)+2(k+1) }{ 2 } \)
\(=\cfrac { (k+1)(k+2) }{ 2 } \) 
\(\therefore P(k+1)\) is true
Thus if P(k) is true, then P(k + 1) is also true
\(\therefore P(n)\) is true for all \(n \in N\) 
Hence \(1+2+3+...+n=\cfrac { n(n+1) }{ 2 } ,n\in N\)

Given equations are
x - y + 2z = 3, 2x + z = 1 and 3x + 2y +z = 4.
The given equations can be written in matrix form as 
\(\begin{bmatrix} 1&-1&2\\2&0&1\\3&2&1 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 3\\1\\4 \end{bmatrix}\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{l} 3 \\ 1 \\ 4 \end{array}\right)\)
\(|A|=1(0-2)+1(2-3)+2(4-0)\)
\(=-2-1+8=5 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 1 & 4 \\ 5 & -5 & -5 \\ -1 & 3 & 2 \end{array}\right)\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} \mathrm{A}=\frac{1}{5}\left(\begin{array}{ccc} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \end{array}\right)\)
\(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
\(={{1}\over{5}}\begin{bmatrix} -2&5&-1\\1&-5&3\\4&-5&2 \end{bmatrix}\begin{bmatrix} 3\\1\\4 \end{bmatrix}={{1}\over{5}}\begin{bmatrix} -6+5-4\\3-5+12\\12-5+8 \end{bmatrix}={{1}\over{5}}\begin{bmatrix} -5\\10\\15 \end{bmatrix}=\begin{bmatrix} -1\\2\\3 \end{bmatrix}\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right)\)
\(x=-1, y=2, z=3\)

a₁₁ = 30, a₁₂ = 40, x₁ = 120
a₂₁ = 20, a₂₂ = 10, x₂ = 60
\({b}_{11}={{{a}_{11}}\over{x_1}}={{30}\over{120}}={{1}\over{4}}\)
\({b}_{12}={{{a}_{12}}\over{{x}_{2}}}={{40}\over{60}}={{2}\over{3}}\)
\({b}_{21}={{{a}_{21}}\over{x_1}}={{20}\over{120}}={{1}\over{6}}\)
\({b}_{22}={{{a}_{22}}\over{{x}_{1}}}={{10}\over{60}}={{1}\over{6}}\)
The technology matrix is   B =  \(\begin{bmatrix}{{1}\over{4}}&{{2}\over{3}}\\ {{1}\over{6}}&{{1}\over{6}} \end{bmatrix}\)
I - B = \(\begin{bmatrix} 0&0\\0&1 \end{bmatrix}-\begin{bmatrix} {{1}\over{4}} &{{2}\over{3}}\\{{1}\over{6}}&{{1}\over{6}} \end{bmatrix}=\begin{bmatrix} {{3}\over{4}}&{{-2}\over{3}}\\ {{-1}\over{6}}&{{5}\over{6}} \end{bmatrix}\)
\(|I-B|=\frac{3}{4} \times \frac{5}{6}-\frac{2}{3} \times \frac{1}{6}=\frac{5}{8}-\frac{1}{9}=\frac{37}{72}=0\)
Since diagonals of I - B are positive and | I - B | is positive, the system is viable
\({(I-B)}^{-1}={{1}\over{|I-B|}}adj\ (I-B)={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\)
X = (I - B)^-1 D where D = \(\begin{bmatrix} 80\\40 \end{bmatrix}\)
\(={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\begin{bmatrix} 80 \\ 40 \end{bmatrix}\)
\(=\frac{72}{37}\left(\begin{array}{cc} 66.67 & +26.67 \\ 13.33 & +30 \end{array}\right)=\frac{72}{37}\left(\begin{array}{l} 93.34 \\ 43.33 \end{array}\right)\)
\(=\left(\begin{array}{c} 181.63 \\ 84.32 \end{array}\right)\)
The output for production section X and Y are 181.63 and 84.32 respectively.

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

\(\frac { 1 }{ { x }^{ 2 }-1 } =\frac { 1 }{ (x+1)(x-1) } =\frac { A }{ x+1 } +\frac { B }{ x-1 } \)
⇒ \(\frac{1}{(x+1)(x-1)}=\frac{A(x-1)+B(x+1)}{(x+1)(x-1)}\)
⇒ 1 = A (x - 1) + B (x + 1)       ...(1)
If x = -1 in (1) we get
1 = A(-2) + 0 ⇒ A = \(\frac { -1 }{ 2 } \)
If x = 1 in (1) we get,
1 = 0 + B(1 + 1) ⇒ 1 = 2B ⇒ B = \(\frac { 1 }{ 2 } \)
\(\frac{1}{x^2-1}=\frac{-1}{2(x+1)}+\frac{1}{2(x-1)}\).

\(28 C_{2 r}: 24 C_{2 r-4}=225: 11\)
\(\frac{28 C_{2 r}}{24 C_{2 r-4}}=\frac{225}{11}\)
\(\frac{\frac{28 !}{(2 r) !(28-2 r) !}}{\frac{24 !}{(2 r-4) !(24-2 r+4) !}}=\frac{225}{11}\)
\(\frac{28 !(2 r-4) !}{(2 r) !(24) !}=\frac{225}{11}\)
\(\frac{28 \times 27 \times 26 \times 25 \times 24 !(2 r-4) !}{(2 r)(2 r-1)(2 r-2)(2 r-3)(2 r-4) ! 24 !}=\frac{225}{11}\)
\((2 r)(2 r-1)(2 r-2)(2 r-3) =28 \times 3 \times 26 \times 11 \)
\(=14 \times 2 \times 3 \times 13 \times 2 \times 11 \)
\((2 r)(2 r-1)(2 r-2)(2 r-3) \)
\(=14 \times 13 \times 12 \times 11 \)
\(2 r =14 \)
\(r =7 \)

\(n P_r=1680 ; n C_r=70\)
\(\frac{n P_r}{r !}=70\)
\(\frac{1680}{r !}=70\)
\(r !=\frac{1680}{70}=24\)
\(r !=4 ! \Rightarrow r=4\)
\(n P_4=1680\)
\(n(n-1)(n-2)(n-3)=8 \times 7 \times 6 \times 5\)
\(n=8\)

(n + 2)! = 60(n - 1)!
(n + 2)(n + 1)(n)(n - 1)! = 60(n - 1)!
(n + 2)(n + 1)(n) = 60
(n + 2)(n + 1)(n) = 5 \(\times \) 4  \(\times \) 3
n = 3
(Counting 60 as product of 3 consecutive natural numbers)

LHS = \(\begin{vmatrix}0 &ab^2 &ac^2 \\a^2b & 0 & bc^2\\a^2c&b^2c&0\end{vmatrix}\)
Taking a, b and c common from R₁, R₂, R₃
\(=\operatorname{abc}\left|\begin{array}{ccc} 0 & b^2 & c^2 \\ a^2 & 0 & c^2 \\ a^2 & b^2 & 0 \end{array}\right|\)
Taking a², b², c² common from C₁, C₂, C₃
= \(a^2b^2c^2\begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 0 \end{vmatrix}\)
= a³b³c³ [0 - 1 (0 - 1)+ 1(1 - 0)]
= a³b³c³ (1 + 1) = 2 a³b³c³ = RHS

\(AB=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}\begin{bmatrix} 3&0\\1&-2 \end{bmatrix}=\begin{bmatrix} 9-1&0+2\\6+1&0-2 \end{bmatrix}=\begin{bmatrix} 8&2\\7&-2 \end{bmatrix}\)
= -16 - 14 = -30
\(\therefore\) |AB| = -30

Let B = \(\begin{vmatrix} 1 &-3 &2 \\4 &-1&2\\3&5&2 \end{vmatrix}\)
Minor of 1 = M₁₁ = \(\begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix}=-2-10=-12\)
Minor of -3 = M₁₂ = \(\begin{vmatrix}4 &2 \\ 3 & 2 \end{vmatrix}=8-6=2\)
Minor of 2 = M₁₃ = \(\begin{vmatrix} 4 & -1 \\ 3 & 5\end{vmatrix}=20+3=23\)
Minor of 4 = M₂₁ = \(\begin{vmatrix} -3 & 2 \\5 & 2 \end{vmatrix}=-6+10=-16\)
Minor of -1 = M₂₂ = \(\begin{vmatrix}1 & 2 \\ 3 & 2 \end{vmatrix}=2-6=-4\)
Minor of 2 = M₂₃ = \(\begin{vmatrix} 1& -3 \\3 &5 \end{vmatrix}=5+9=14\)
Minor of 3 = M₃₁ = \(\begin{vmatrix} -3 &2 \\ -1 & 2 \end{vmatrix}=-6+2=-4\)
Minor of 3 = M₃₂ = \(\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=2-8=-6\)
Minor of 2 = M₃₃ = \(\begin{vmatrix} 1 & -3 \\4 & -1 \end{vmatrix}=-1+12=11\)
Co-factor of 1 = A₁₁ = (-1)¹⁺¹ M₁₁ = -12
Co-factor of -3 = A₁₂= (-1)¹⁺² M₁₂ = -2
Co-factor of 2 = A₁₃= (-1)¹⁺³ M₁₃ = 23
Co-factor of 4 = A₂₁ = (-1)²⁺¹ M₂₁ = 16
Co-factor of -1 = A₂₂ = (-1)²⁺² M₂₂ = -4
Co-factor of 2 = A₂₃ = (-1)2+3 M₂₃ = -14
Co-factor of 3 = A₃₁ = (-1)³⁺¹ M₃₁ = -4
Co- factor of 5 = A₃₂ = (-1)³⁺² M₃₂ = 6
Co-factor of 2 = A₃₃ = (-1)³⁺³ M₃₃ = 11

\(x^{2}[8-1]-x[0-3]+1[0-6]=28 \)
\(7 x^{2}+3 x-6-28=0 \)
\(7 x^{2}+3 x-34=0 \)
\(7 x^{2}+17 x-14 x-34=0 \)
\(x(7 x+17)-2(7 x+17)=0 \)
\((x-2)(7 x+17)=0 \)
Integral value of x is 2

  \(\cfrac { 1 }{ 9! } +\cfrac { 1 }{ 10! } =\cfrac { n }{ 11! } \)
\(\cfrac { 1 }{ 9! } +\cfrac { 1 }{ 10\times 9! } =\cfrac { n }{ 11! }\)
\( \cfrac { 1 }{ 9! } \left[ 1+\cfrac { 1 }{ 10 } \right] =\cfrac { n }{ 11! } \)
\( \cfrac { 1 }{ 9! } \times \cfrac { 11 }{ 10 } =\cfrac { n }{ 11! } \)
\(n=\cfrac { 11!\times 11 }{ 9!\times 10 } \)
\(=\cfrac { 11!\times 11 }{ 10! } \)
\( =\cfrac { 11\times 10!\times 11 }{ 10! } \)
\(n=121\)

Let A = \(\left[ \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right| \)
= 4 – 4 = 0
\(\therefore\) A is a singular matrix

\(\left| \begin{matrix} 1 & 3 & 4 \\ 102 & 18 & 36 \\ 17 & 3 & 6 \end{matrix} \right| =6\left| \begin{matrix} 1 & 3 & 4 \\ 17 & 3 & 6 \\ 17 & 3 & 6 \end{matrix} \right| \)
= 0 (since R₂ ≡ R₃)

The letter of the word CHAT in alphabetical order are A, C, H, T.
(i) Number of words starting with A = 3! = 6 C begins
(ii) Number of words starting with CA are 2! = 2
Now the words CH begins
(iii) After that we get the word CHAT = 1
\(\therefore\) Rank of CHAT is = 6 + 2 + 1 = 9

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
\(=(n-1) !=(8-1) !=7 !\)