A car can be bought by choosing a brand, then a variant model, and then a colour. A brand can be chosen in 2 ways; a model can be chosen in 3 ways and a colour can be chosen in 5 ways. By the rule of product, the person can buy a car in 2 \(\times\) 3 \(\times\)5 = 30 different ways.

\({ a }_{ n }=\begin{cases} n+1\quad if\quad n\quad is\quad odd \\ n\quad \quad if\quad n\quad is\quad even \end{cases}\)
a₁ = 1 + 1 = 2, a2 = 2, a₃ = 3 + 1 = 4
a₄ = 4, a₅ = 5 +1 = 6, a₆ = 6
hence the first 6 terms are 4, 2, 2, 4, 6, 6...

n elements.
Number of subsets with no element = nC₀
Number of subsets with 1, 2, 3,4. elements are nC_o + nC₁ + nC₂ + ... + nC_n respectively
∴ Total number of subsets
= nC₀+nC₁+nC₂+...+nC_n
= Sum of the co-efficient in the binomial expansion (x + a)ⁿ
= 2ⁿ

5 cars can be chosen out of 100 cars in ¹⁰⁰C₅ ways
=​​​​
= 451725120

There are 6 letters in the word 'SIMPLE'.
So, total number of words is equal to the number of arrangements of these letters, taken all at a time. Sum order of such arrangements is 6 P₆ = 6! = 720

= 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 + 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1
= 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 (1+5)
= 24 \(\times\) 6 = 144

To arrange 5 persons in a row, we need 5 place.
Number of ways of 1st person can be seated in a row = 5
Number of ways of 2nd person can be seated in a row = 4
Number of ways of 3rd person can be seated in a row = 3
Number of ways of 4th person can be seated in a row = 2
Number of ways of 5th person can be seated in a row = 1
∴ Number of ways of 5 persons' can be seated in a row = 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 5!
= 120

There are 11 letters in the word, with respect to number of repetitions of letters there are 4 distinct letters (R, S, T, N), 2 sets of two alike letters (PP, II), 1 set of three alike letters (OOO). The following table will illustrate the combination of these sets and the number of words.

Hence, the total number of strings are 2520 + 20 + 300 + 450 + 3600 = 6890.

a = k^x , b = k^y and c = k^z
Also, given that a,b, c are in GP
\(\frac { b }{ a } =\frac { c }{ b } \Rightarrow { b }^{ 2 }=ac\)
ky² = k^x .k^z
k²y = k^x+z
2y  = x +z
y + y = x + z
y - x = z - y
Common difference is same for x, y, z
∴ x, y, z are in arithmetic progression

(i) Find the linear relationship between C and F.
By the given data
x₁ (100°C)  y₁(212°F)
x₂(0°C) y₂(32°F)
Using two point form, the linear relationship between C and F is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac { y-212 }{ 32-212 } =\frac { x-100 }{ 0-100 } \)
\(\Rightarrow \quad \frac { y-212 }{ -180 } =\frac { x-100 }{ -100 } \)
\(\Rightarrow \quad \frac { y-212 }{ 9 } =\frac { x-100 }{ 5 } =\frac { 5 }{ 9 } (y-212)=x-100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-212)+100\quad \Rightarrow \quad x=\frac { 5 }{ 9 } y-\frac { 5 }{ 9 } \times 212+100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } y-118+100\quad \Rightarrow x=\frac { 5 }{ 9 } y-18\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-32) \Rightarrow C=\frac { 5 }{ 9 } (F-32)\quad .....(1)\)
[\(\because \)  x represents Celsius and y represents Fahrenheit]
Which is the required relationship between C and F.
(ii) Find the value of C for 98.6°F and
Find C when F = 98.6° F
Substituting F = 98.6° in (1) we get,
C = \(\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=\frac{333}{9}=37°\)
(iii) Find the value of F for 38°C.
Substituting C = 38° in (1) we get,
38=\(\frac{5}{9}(F-32)\)
⇒ \(\frac{342}{5}+32\) = F
⇒ F = 100.4°C

To form a triangle we need 3 points.
Considering 7 points as group A and the remaining 8 points as Group B, the following choices are possible

∴ Required number of ways of forming the triangle
= \({ 7C }_{ 2 }\times { 8C }_{ 1 }+{ 7C }_{ 1 }+{ 8C }_{ 2 }\)
= \(\frac { 7\times 6 }{ 2\times 1 } \times 8+\frac { 7\times 8\times 7 }{ 2\times 1 } \)
= \(21\times 8+7\times 4\times 7\)
= 168 + 196
= 364

To form a triangle we need minimum 3 non-collinear points.
points can be selected from 15 non-collinear points in 15C₃ ways.
= \(\frac { 15\times 14\times 13 }{ 3\times 2\times 1 } \)
= 455

Four subjects can be arranged on the shelf in 4! ways.
The books on mathematics can be arranged in 4! ways, physics on 3! ways, chemistry on 2! ways and Biology on 1! ways.
Hence, total number of ways of arranging the books
= 4! 4! 3! 2! 1!
= \((4\times 3\times 2\times 1)(4\times 3\times 2\times 1)(3\times 2)(2\times 1)\)
= (24)(24)(6)(2)
= 6912

Let p( n) be the statement \(1^2+2^2+3^2+L+n^2>{n^3\over 3}\)
Step1: Putting n = 1, we get
\(1>{1^3\over 3}⇒1>{1\over 3}\) Which is true
∴ p(l) is true.
Step 2:
Let us Assume that p(k) is true.
∴ \(1^2+2^2+3^2+...+k^2>{k^3\over 3}\)
Step 3:
To prove that p( k + 1) is true.
ie to P.T.  \(1^2+2^2+3^2+ ... k^2+(k+1)^2>{(k+1)^3\over3}\)
Consider 1² + 2² + 3² + ... k² + (k + 1)²
\(>{K^3\over 3}+(k+1)^2\) [using (1)]
\(> {k^3\over 3}+k^2+2k+1\)
\(>{1\over 3}[k^2 + 3k^2 + 6k + 3]\)
[∴ (a + b)³ = a³+ 3a² b + 3 ab² + b³]
\(>{1\over 3}{k+1}^3\)
∴ p(k + 1) is true
Hence, by the principle of mathematical induction,p(n) is true for all values of m.

Given a = 5
Since the particle gets doubled, the G.P will be 5, 10,20,40, ... 1,50,000
⇒ a.r^n-1 > 1,50,000
⇒ a.(2^n-1) > 1,50,000
⇒ 5(2^n-1) > 1,50,000
⇒ \(2^{n-1}>{1,50,000\over5}\)
⇒ 2^n-1 > 30,000
⇒ 2^n-1 > 2⁴ x 1875
⇒ \({2^{n-1}\over24}>1875\)
⇒ 2^n-5 > 1875
⇒ (n - 5) log 2 > log 1875
⇒ \(n-5>{log1875\over log2}\)
⇒ \(n-5>{3.2730\over 0.3010}\)
⇒ n-5 > 10.873
⇒ n > 10.873 + 5
⇒ n > 15.873
⇒ n = 15
Hence the 15^th day, the infectious Virus particles just grow over 1,50,000 units.

Let P(n) be the statement \(\left(1-{1\over 2^2} \right)\left(1-{1\over 3^2} \right)\left(1-{1\over 4^2} \right)...\left(1-{1\over n^2} \right)={n+1\over 2n}\)
Step 1:
Putting n = 2, we get
\(\left(1-{1\over2^2}\right)\left(1-{1\over3^2}\right)={2+2\over 2(2)+2}\)
\(⇒\ \left(1-{1\over4}\right)\left(1-{1\over9}\right)={4\over 6}\)
\(⇒\ \left(3\over4\right)\left(8\over9\right)={2\over3}⇒{2\over3}={2\over3}\)
∵ p(1) is true
Step 2:
Let we assume that p(k) is true
\(∵\ \left(1-{1\over 2^2}\right)\left(1-{1\over 2^2}\right)...\left(1-{1\over (K+1^2)}\right)={K+2\over 2K+2}\)
Step 3:
To prove that p(K+1) is true

i.e to P.T \(\left(1-{1\over2^2}\right)\left(1-{1\over 3^2}\right)...\left(1-{1\over (K+1)^2}\right)\left(1-{1\over (K+2)^2}\right)={K+3\over 2(K+1)+2}={k+3\over 2K+4}\)
\(LHS=\left(1-{1\over 2^2}\right)\left(1-{1\over 3^2}\right)...\left(1-{}1\over (K+1)^2\right)\left(1-{1\over (K+2)^2}\right)\)
\(={K+2\over 2K+2}\left(1-{1\over (K+2)^2}\right)\)
\({K+2\over 2K+2}\left({(K+2)^2-1\over (KK+2)^2}\right)\) [Using (1)]
\(={K+2\over 2K+2}\left(K^2+4+4K-1\over (K+2)^2\right)\)
\(={K+2\over 2K+2}\left(K^2+4K+3\over (K+2)^2\right)={K+2\over 2(K+1)}{{(K+1)(K+3)\over (K+2)^2}}\)
\(={K+3\over 2(K+2)}={K+3\over 2K+4}=RHS\)
∵ p(K+1) is true
Hence by mathematical induction, p(n) is true for all values of n.

\((2+2 x)^{10} \text { Term containing } x^{6} \text { is }\)
\({ }^{10} \mathrm{C}_{6}(2)^{10-6}(2 x)^{6}={ }^{10} \mathrm{C}_{6} 2^{4} 2^{6} x^{6}\)
\(\text { Coefficient of } x^{6} \text { is }{ }^{10} \mathrm{C}_{6} 2^{10}\)

Perpendicular distance from vertex to the opposite side is \(\frac{2+3-2}{\sqrt{1+1}}=\frac{3}{\sqrt{2}}\)
\(\frac{3}{\sqrt{2}}=\frac{\sqrt{3}}{2} a\)
\(\Rightarrow a =\frac{6}{\sqrt{6}} \)
\(\therefore a =\sqrt{6}\)

\(\text { Given }{ }^{n} C_{4},{ }^{n} C_{5},{ }^{n} C_{6} \text { are in A.P }\)
\({ }^{2 n} \mathrm{C}_{5}={ }^{n} \mathrm{C}_{4}+{ }^{n} \mathrm{C}_{6}\)
\(\frac{2\lfloor n}{\lfloor n-5\lfloor 5}=\frac{\lfloor n}{\lfloor n -4\lfloor 4}+\frac{n}{\lfloor n -6\lfloor 6}\)
\(\frac{2}{\lfloor n-5\lfloor 5} =\frac{1}{\operatorname{\lfloor n}-4\lfloor 4}+\frac{1}{\lfloor n-6\lfloor 6}\)
\(\frac{2(n-4) 6}{(n-4)\lfloor n-5\lfloor 5.6}=\frac{5.6}{\lfloor-45.6\lfloor 4}+ \frac{(n-4)(n-5)}{\lfloor 6(n-4)(n-5) \lfloor n-6}\)
\(\Rightarrow \frac{12(n-4)}{\lfloor n-4\lfloor 6}=\frac{30}{\lfloor n-4\lfloor 6}+ \frac{(n-4)(n-5)}{\lfloor n-4 \lfloor 6}\)
\(12 n-48 =30+n^{2}-9 n+20 \)
\(n^{2}-21 n+98 =0 \)
\((n-14)(n-7) =0 \)
\(n=14(\text { or }) n =7 \)

\(\text { Slope of } 3 x-y+5=0 \text { is } \frac{-3}{-1}=3=m_{1}\)
Let m₂ be the slope of the second line
\(\text {Given } \tan \theta=\tan 45^{\circ}=1 \Rightarrow \frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=\pm 1\)
\(\frac{3-m_{2}}{1+3 m_{2}}=1 \quad \frac{m_{2}-3}{1+3 m_{2}}=1\)
\(3-m_{2}=1+3 m_{2} \quad m_{2}-3=1+3 m_{2}\)
\(2=4 m_{2} \quad-2 m_{2}=4\)
\(\mathrm{m}_{2}=\frac{1}{2} \quad \mathrm{~m}_{2}=-2\)
\(\left(\frac{1}{2},-2\right)\)

\(\text { Number of triangles }={ }^{10} \mathrm{C}_{3}-{ }^{4} \mathrm{C}_{3}\)
\(=\frac{10 \times 9 \times 8}{1 \times 2 \times 3}-4 \)
\(=120-4 =116 \)

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

\(\text { No. of lines }{ }^{10} \mathrm{C}_{2}-{ }^{4} \mathrm{C}_{2}+1=45-6+1=40\)

zero

Each of the 10 bulbs are operated independently means that each bulb can be operated in two ways. That is in off mode or on mode. The total number of doing this are 2¹⁰ which includes the case in which 10 bulbs are off. Keeping all 10 bulbs in "off" mode, the room cannot be illuminated. Hence, the total number of ways are 2¹⁰ - 1 = 1024 - 1 = 1023.

The following figure illustrates the electricity distribution network.

There are 110 smaller capacity transformer attached to a larger capacity transformer. As each smaller capacity transformer can be linked with only 100 consumers, we have for the 109 transformers, there will be 109 \(\times\)100 = 10900 links. For the 110^th transformer, there are only 29 consumers linked. Hence, the total number of consumer linked to the 238^th larger capacity transformer is 10900 + 29 = 10929.

Given \(\alpha\) = 30° and p = 12
Equation of the straight line in normal form is x cos\(\alpha\)+y sin \(\alpha\) = p
⇒ x cos 30°+ y sin 30° = 12

⇒ \(x(\frac{\sqrt{3}}{2})+y(\frac{1}{2})=12\)
\(\frac{\sqrt{3}x+y}{2}=12\)
√3x + y = 24

(i) Two particular books are always selected
Two particular books are always selected, the remaining 3 books can be selected from 10 books in 10C₃ ways.

(ii) Two particular books are never selected
Since two books are never to be selected, the selection of 5 books from 10 books are done in 10C₅ ways
\(\frac { 10! }{ 5!5! } =\frac { 10\times 9\times 8\times 7\times 6\times 5\times ! }{ 5!5\times 4\times 3\times 2\times 1 } \)
= \(\frac { 10\times 9\times 8\times 7\times 6\times }{ 5\times 4\times 3\times 2 } \)
= 252