Given p(n) : " 10n + 3" is prime
\(\therefore\) p(2) = 10(2) + 3 = 20 + 3 = 23 which is a prime number.
\(\Rightarrow\) p(2) is true.
Now, p(3) = 10(3) + 3 = 30 + 3 = 33 which is not a prime number.
\(\Rightarrow\) p(3) is not true.

The total number of ways of selecting 11 players out of 15 players.
= 15C₁₁
= 15 C_15-11 = 15C₄   [nC_r = nC_n-r]
\(=\frac { 15\times 14\times 13\times \times 12 }{ 4\times 3\times 2\times 1 } =1365\)

=  \(\frac { (n+3)(n+2)(n+1)! }{ (n+1)! } \) 
= (n + 3) (n + 2)
= n² + 3n + 2n + 6 = n²+ 5n + 6

The total number of signals is equal to the number of ways of filling 3 places in succession by 4 flags of different colours.
The upper place can be filled in 4 ways, following which the next place can be filled in 3 ways and the lower place can be filled in 2 ways.
Hence, by fundamental principle of multiplication, the required number of signals = 4 \(\times\) 3 \(\times\) 2 = 24.

Let p(n) be the statement 41n - 1411 is a multiple of 27.
Step 1: p(1): 41¹- 14¹ is a multiple of 27.
⇒ p(1): 27 is a multiple of27.
⇒ p(1) is true.
Step 2: Let p(m) be true.
Then 41^m - 14^m is a multiple of 27.
⇒ 41^m -14^m=⋋.27
Step 3: To prove that p(m + 1) is true.
i.e. to prove that 41^m+1- 14^m+1 is a multiple of 27
Consider 41^m+1- 14^m+1
41^m+1 - 41 x 14^m + 41 x 14^m - 14^m+1    [Adding and subtracting 41 x 14^m]
= 41^m . 41 - 41.14^m + 41 x 14^m - 14^m. 14
= 41(41^m -14^m) + 14^m (41 - 14)
= 41 (⋋.27) + 14^m(27)
= 27 (41⋋ + 14^m) which is a multiple of27
⇒ p(m + I) is true. ,
Hence, by the principle of mathematical induction,p(n) is true for all n E N.

The plus signs can be arranged in only one way, because all are identical, as shown below.

A blank box in the above arrangement shows the available space for the minus sign.
The 5 minus signs are now to be arranged in 8 boxes.
So, 5 boxes out of 8 boxes can be chosen in 8C₅ ways.
Since all minus signs are identical, so 5 minus signs can be arranged in 5 chosen boxes in only one way.
Hence, the number of possible arrangements
1\(\times\)8C₅\(\times\)1
= \(\frac{8!}{5!3!}=\frac{8\times7\times6\times5!}{3\times2}\) = 56

Given (n + 1)C₈ : (n -3) P₄ = 57 : 16
⇒ \(\frac { (n+1){ C }_{ 8 } }{ (n-3){ P }_{ 4 } } =\frac { 57 }{ 16 } \)
⇒ 16(n+2)C₈ = 57(n-3)P₄
\(\frac { 16(n+1)! }{ 8!(n+1-8)! } =\frac { 57(n-3)! }{ (n-3-4)! } \)      \(\left[ \because nP_{ r }=\frac { n! }{ (n-r)! } ,n{ C }_{ r }=\frac { n! }{ r!(n-r)! } \right] \)

(n+1) n(n-1) (n-2) = \(\frac { 57\times 8! }{ 16 } \)
=
= \(3 \times 19 \times 7 \times 6 \times 54 \times 3\)
\((n+1) n(n-1)(n-2)=21 \times 20 \times 19 \times 18\)
n = 20

(i) Either starts with L or ends with S.

Since the words starts with L, the remaining 4 boxes can be filled in 4 x 3 x 2 x 1 ways by the remaining letters 0, T, U, S.
∴ Number of words starting with L
= 1 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24.

Here also, the remaining 4 boxes can be filled in 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 ways = 24.......(1)
Number of words ending with S = 24 ....(2)
Number of words starting with L and end with S are 3 \(\times\) 2 \(\times\) 1 = 6...(3)
∴  By fundamental principle of addition, number of words either starts with L nor ends with S = 24 + 24 - 6 = 48 - 6 = 42

(ii) Neither starts with L nor ends with S.
Total number of words formed by the letters of the word LOTUS is 5 \(\times\) 4 \(\times\) 3 \(\times\) 2\(\times\) 1 = 120.
Now, number of words neither starts with L nor end with S.
= (Total number of words) - (Number of words starts with either L nor ends with S)
= 120 - 42
= 78.

(i) Repetition of digits are not allowed?

Unit digit can be filled in 2 ways using the digit 0 or 5, since the three-digit number is divisible by 5.
Hundreds place can be filled in 4 ways (excluding 0 and 5)
Tens place can also be filled in 4 ways by the remaining digits.
∴ By fundamental principle of multiplication, required number of 3 digit numbers = 4 \(\times\) 4 \(\times\) 2 = 32 Ways.
(ii) Repetition of digits are allowed?

Unit place can be filled in 2 ways using the digit 0 and 5, since the three digit number is divisible by 5.
Hundreds place can be filled in 5 ways excluding 0
Tens place can be filled in 6 ways, since repetition of digits are allowed.
∴ By fundamental principle of multiplication, required number of three digit numbers = 5 \(\times\) 6 \(\times\) 2 = 60 ways

The repetition of digits is allowed

The unit place can be filled in 3 ways using the digits 1, 3, or 5 since we need 3 digit odd numeric Hundreds place can be filled in 5 ways excluding 0 and repetition of digits is allowed.
Tens place can be filled in 6 ways .
∴ By fundamental principle of multiplication, required number of 3 = digit odd numbers
= 5 \(\times\) 6 \(\times\) 3 = 30 \(\times\) 3 = 90.

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

The number of five digit telephone numbers which can be formed using the digits 0,1,2.... 9 is 10⁵.
The number of 5 digit numbers which has none of their digit repeated is ¹⁰P₅, = 30240
The required number of telephone. number is 10⁵ =- 30240 = 69,760

No. of ways of answering = 5³ = 125
Correct answer - 1
.'. Number of incorrect answer = 125 - 1 = 124

p(n) 2³ⁿ - 1 is divisible by 8
For n = 1, we get
P(1) = 3^2.1-1 = 9 - 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k)3^2k - 1 is divisible by 8......(1)
Now,  P(K+1) = 3^{(2k + 2)} - 1 = 3^2k.3² - 1
= 3² (3^2k - 1) + 8
Now,3^2k-1 is divisible by 9. [Using (1)]
\(\therefore\) 3² (3^2k - 1) + 8 is also divisible by 8.
Hence, 3²ⁿ - 1 is divisible by 8\(\forall\) n \(\in\) N

Total number of points = 18
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
\(\therefore\) Total number of straight line formed by joining 2 points out of 18 points = ¹⁸C₂
Number of straight lines formed by joining 2 points out of 5 points = ⁵C₂
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are
=¹⁸C₂ -⁵C₂ +1 = \({18 ·17\over2.1}-{5·4\over2.1}+1= 153 -10 + 1-144\)
Hence, the total number of straight lines = 144

The given strings contains 6 letters (V, O, W, E, L, S).
(i) Since all strings must begin with E, we have the remaining 5 letters which can be arranged in ⁵P₅ = 5! ways.
Therefore the total number of strings with E as the starting letter is 5! =120.

(ii) Since all strings must begin with E, and end with W, we need to fix E and W. The remaining 4 letters can be arranged in ⁴P₄ = 4! Ways.

Therefore the total number of strings with E as the starting letter and W as the final letter is 4! = 24.

There are 9 letters in the word ALLAHABAD.
(i) Out of which 4 are A's, 2 are L's and the rest are all distinct.
So, the required number of words = \(\frac { 9! }{ 4!2! } \)
\(=\frac { 9\times 8\times \times 7\times \times 6\times 5\times 4! }{ 4!\times 2! } \)
= 9 \(\times\)4\(\times\)7\(\times\)6\(\times\)5 = 7560.
(ii) There are 4 vowels and all are alike.
Since there are 4 even places, 4 vowels can occupy the even Places in \(\frac { 4! }{ 4! } =1\)  way.
Now, we are left with 5 places and 5 letters out of which 2 are L's and other distinct.
\(\therefore\) These can be arranged in \(\frac { 5! }{ 2! } \) ways.
\(\therefore\) Total number of words in which vowels occupy the even places = \(\frac { 5! }{ 2! } \times \frac { 4! }{ 4! } =\frac { 5\times 4\times 3\times \times 2\times 1 }{ 2\times 1 } \times 1\)
= 60 ways.