When atleast 3 girls are included, then
Number of ways = ⁴C₃ \(\times\)⁷C₂ + ⁴C₄ \(\times\)⁷C_I
\(=4\times {7\times6\over2\times1}+1\times7=84+7=91ways\)
Hence the required number of ways are 91 ways

There are 6 periods in each working day of a school.
Since each subject is allowed at least one period, we first select one subject for the left out period. This can be done in 5C₁ ways.
Now, six subject can be arranged in \(=\frac { 6! }{ 2! } \) ways
Hence, total number of permutations = 5C₁ \(\times\)\(\frac { 6! }{ 2! } \)ways

= 1800.

We are given 8 letters namely AAABBCCC.
Clearly three are of one kind, 2 are of second kind and C are of third kind.
\(\therefore\) Total number of permutations = \(\frac { 8! }{ 3!2!3! } \)

= 560.

Let us keep EN together and consider it as one letter.
Now, we have 5 letters which can be arranged in a row in 5P₅ = 5! = 120 ways.
Hence, the total number of ways in which N is always next to E is 120.

Since the passcode has 6 distinct digits, the first digit can be tried in 10 ways using 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Similarly 2^nd, 3^rd, 4^th, 5^th and 6^th digit can also be tried in 9, 8, 7, 6, 5, 10 respectively.
∴ Maximum number of attempts made to retrieve the passcode = 10 \(\times\) 9 \(\times\) 8 \(\times\) 7 \(\times\) 6 \(\times\) 5
= 151200.

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

Given,  ⁵p_r= 2⁶p_r-1
\(\Rightarrow {5!\over (5-r)!}={6!\over \{6-(r-1)\}!}\)
\(\Rightarrow {5!\over (5-r)!}={6.5!\over (7-r)!}\)
\(\Rightarrow {5!\over (5-r)!}={6\over (7-r)(6-r)(5-r)!}\)
\(\Rightarrow 1={6\over (7-r)(6-r)} \Rightarrow (7-r)(6-r)=6\)
\(\Rightarrow\) 42 - 13r + r² = 6 \(\Rightarrow\)r² - 13r + 36 = 0
\(\Rightarrow\) (r-4)(r-9)=0\(\Rightarrow\) r =4,9
Now, we know that ⁿP_r is meaningful only when r \(\le\) n.
\(\therefore\)⁵p_rand ⁶p_r-1are meaningless when r = 9.
\(\therefore\)Rejecting r = 9, we have r = 4​​​​​​​

Given nP_r = nP_{r+ 1}

⇒ 1 = n-r       ...(1)
Also, it is given that nC_r = nC_r-1

\(⇒\ {1\over r}={1\over n-r+1}\)
⇒ n - r + 1 = r
⇒ n -2r = -1 ...(2)
(1) - (2} ⟶ 2 = r
Substituting r = 2 in (1) we get,
n-2 = 1 ⇒ = 3
∴ n = 3 and r = 2
n = 3 and r = 2

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) Repetition of digit is allowed

Since we are going to find numbers between 100 and 500 it has 3 = digits
The unit place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5
The tens place also can be filled in 6 ways since repetition of digits is allowed.
The hundreds place can be filled in 4 ways using the digits 1, 2, 3, 4 [excluding 0 and 5]
∴ By fundamental principle of multiplication, required number of 3 - digit numbers = 4 \(\times\) 6 \(\times\) 6 = 144.
(ii) Repetition of digits is not allowed.

Hundreds place can be filled in 4 ways excluding 0 and 5
Tens place can be filled in 5 ways since repetition of digits is not allowed
Unit place can be filled in 4 ways.
∴ By fundamental principle of multiplication, required number of three-digit numbers = 4 \(\times\) 5 \(\times\) 4 = 80.

\(a^{2}-a^{a} C_{2} =a^{2}-a^{a} C_{4} \)
\(a^{2}-a^{a} C_{2} =a^{2}-a^{a} C_{a-a-4}\left(\because^{n} C_{r}={ }^{n} C_{n-r}\right) \)
\(a^{2}-a-4 =2 \)
\(a^{2}-a-6 =0 \)
\((a-3)(a+2) =0 \)
\(a=3 \text { or } a=-2 \text { which is impossible }\)

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

Total numbers = 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24,
Sum of alt integers in tenth place
= 6 ( 2 + 4 + 5 + 7) = 108

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

Given (n+2)! = 60(n-1)!

\(\Rightarrow \) (n+2) (n+1)(n) = 60
\(\Rightarrow \) (n+2)(n+1)(n) = 5 \(\times\)4 \(\times\)3
Equating the terms both sides we get, n = 0

LHS = n!(n + 2)
RHS = n! + (n+1)!
= n!+(n+1)(n)!....(1)
= n!(1+n+1)
= n!(n+20)...(2)
From (1) and (2), LHS = RHS
Hence Proved.

\(\frac { (2n)! }{ n! } =\frac { (2n)(2n-1)(2n-2)...4.3.2.1 }{ n! } \)
\(=\frac { (2n)(2n-2)(2n-4)..4.2(2n-1)(2n-3)....3.1 }{ n!\quad n! } \)   [Separating odd and even terms]
\(=\frac { { 2 }^{ n }.(2n-1)(2n-3)...3.1 }{ n! } \)

Given ¹⁰P_r-1 = 2 \(\times\) ⁶P_r
⇒ \(\frac { 10! }{ (10-r+1)! } =2\times \frac { 6! }{ (6-r)! } \)   \(\left[ \because n{ P }_{ r }=\frac { n! }{ (n-r)! } \right] \)
⇒ \(\frac { 10\times 9\times 8\times 7\times 6! }{ (11-r)! } =\frac { 10\times 9\times 8\times 7\times 6! }{ (11-r)! } \)
⇒ \(\frac { 10\times 9\times 8\times 7 }{ (11-r)(10-r)(8-r)(7-r)(6-r) } =\frac { 2 }{ (6-r)! } \)
⇒ \(\frac { 10\times 9\times 8\times 7 }{ (11-r)(10-r)(8-r)(7-r) } =2\)
\( \Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r)=5 \times 9 \times 8 \times 7 \)
\( \Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r)=7 \times 6 \times 5 \times 4 \times 3 \)
\( \Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r)=(11-4)(10-4)(9-4)(8-7)(7-4) \)
⇒ r = 4