Considering all S as one letter there are 5 letters containing 2 C's, one U, and one E which can be arranged in
\(\frac { 5! }{ 2!1!1! } =\frac { 5\times 4\times 3\times 2 }{ 2 } \)
= 60 Ways

LHS =¹⁵C₃ + 2 \(\times\) ¹⁵C₄ +¹⁵C₄ +¹⁵C₅
= (¹⁵C₃ +¹⁵C₄) + (¹⁵C₄ + ¹⁵C₅)
=(¹⁵C₃ +¹⁵C₄) + (¹⁵C₄ + ¹⁵C₅)       [∴ ⁿC_r-1 + ⁿC_r = n +¹C_r]
= ¹⁶C₄ + (¹⁵C₄ +¹⁵C₅)
=¹⁶C₄ + ¹⁶C₅
= ¹⁷C₅ = RHS.

Given ⁿP_r = 720 and ⁿC_r = 120
⇒ \(\frac { n! }{ (n-r)! } \) = 720 .... (i)
\(\frac { n! }{ r!(n-r)! } \) = 120..... (ii)
⇒ \(\frac { \frac { n! }{ (n-r)! } }{ \frac { n! }{ r!(n-r)! } } =\frac { 720 }{ 120 } \) 
[Dividing (i) by (ii)]
⇒ \(\frac { n! }{ (n-r)! } \times \frac { n! }{ r!(n-r)! } =6\)
⇒ r! = 6 ⇒ r! = 3 \(\times\) 2 \(\times\) 1 = 3!
⇒ r = 3
Substituting r = 3 in (i) we get
\(\frac { n! }{ (n-r)! } =\frac { n! }{ (n-r)! } =720\Rightarrow \frac { n! }{ (n-3)! } =720\)
⇒ \(\frac { n(n-1)(n-2)(n-3)! }{ (n-3)! } \) = 720 ⇒ n(n-1) (n-2) = 720
⇒ n(n-1) (n-2) = 10 \(\times\) 9 \(\times\) 8
⇒ n = 10

In the letters of the word, ARTICLE, there are three vowels namely A, I, E.
There are 3 even places.
3 vowels can occupy the even places in 3P₃ = 3! ways.
Remaining 4 letters can occupy 4 places in 4! ways.
Hence, total number of ways of arrangement = 4! \(\times\) 3!
=  \(4\times 3\times 2\times 3\times 2\)
=144

(i) With repetition

The given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The unit place can be filled in only one way using 3.
Since repetition is allowed, the tens place can be filled in 10 ways using any one of the digits from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The hundreds place can be filled in 9 ways using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (excluding 0)
∴ By fundamental principle of multiplication, total number of 3 digit numbers = 9 \(\times\) 10 \(\times\) 1 = 90.
(ii) Repetition of digits is not allowed

The unit place can be filled in 1 way
Since repetition of digits is not allowed, the tens place can be filled in 8 ways.
Hundreds place can be filled in 8 ways.
∴ Total number of 3-digit numbers without repetition = 1 \(\times\) 8 \(\times\) 8 = 64

(i) The following are the choices to select at least 3 women

∴ Required number of ways of forming the committee
= 8C₄\(\times \)4C₃ + 8C₃\(\times \)4C₄
= \(\frac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2\times 1 } \times 4+\frac { 8\times 7\times 6 }{ 3\times 2\times 1\times } \times 1\)    [∵ 4C₃ = 4C₁ = 4, 4C₄ = 1]
= 280 + 56
= 336
(ii) The following are the choices to select at most 3 women

Hence, required number of ways of forming the 49 committee is
\({ 8C }_{ 4 }\times { 4 }C_{ 3 }+{ 8C }_{ 5 }\times 4{ C }_{ 2 }+{ 8C }_{ 6 }\times 4C_{ 1 }+8C_{ 7 }\times 4C_{ 0 }\)
= \({ 8C }_{ 4 }\times { 4C }_{ 1 }+{ 8C }_{ 3 }\times { 4C }_{ 2 }+{ 8C }_{ 2 }\times { 4C }_{ 1 }+{ 8C }_{ 1 }\times { 4C }_{ 0 }\)
=
= 280 + 336 + 112 + 8 = 736
(iii) The following are the choices to select at most 3 women

Hence, required number of ways of forming the 49 committee is
\({ 8C }_{ 4 }\times { 4 }C_{ 3 }+{ 8C }_{ 5 }\times 4{ C }_{ 2 }+{ 8C }_{ 6 }\times 4C_{ 1 }+8C_{ 7 }\times 4C_{ 0 }\)
=\({ 8C }_{ 4 }\times { 4C }_{ 1 }+{ 8C }_{ 3 }\times { 4C }_{ 2 }+{ 8C }_{ 2 }\times { 4C }_{ 1 }+{ 8C }_{ 1 }\times { 4C }_{ 0 }\)
= 
= 280 + 336 + 112 + 8 = 736

(i) The lexicographic order of the letters of the given word is A, D, E, G, N, R.
The word GARDEN has 6 letters in which no letters are repeating.
∴ Rank \( = (3 \times 5 !)+(0 \times 4 !)+(3 \times 3 !)+ (0 \times 2 !)+(0 \times 1 !)+(0 \times 0 !)+1 = 3 \times 120+0+18+0+0+0+1 \)
 = 360 + 19 = 379
(ii) The lexicographic order of the letters of the given word is A, D, E, G, N, R.
The word GARDEN has 6 letters in 'which no letters are repeating
 Rank = \((1 \times 5 !)+0+(2 \times 3 !)+(1 \times 2 !)\)
 ∴  Rank of DANGER = 135.

Let p(n) be the statement
\(1.2+2.3+...+n(n+1)={n(n+1)(n+2)\over 3}\)
Step 1:
Putting n = 1 we get
\(1.2={1(1+1)(11+2)\over 3}={2(3)\over3}=2⇒2=2\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
∵ \(1.2+2.3+3.4+...+K(K+1)={K(K+1)(K+2)\over3}\)
Step 3: To prove that p(K+1) is true
ie to P.T. \(1.2+2.3+3.4+...+K(K+1)+(K+1)(K+2)={(K+1)(K+2)\over 3}\)
LHS = 1.2 + 2.3 +...+K(K+1)+(K+1)(K+2)
\(={K(K+1)(K+2)\over 2}+(K+1)(K+2)\)
\(=(K+1)(K+2)\left[{K+\over3}+1\right]\)
\(={(K+)(K+2)(K+3)\over 3}=RHS\)
∵ p(K+1) is true.
Hence, by mathematical induction, p(n) is true for all values of n.

(i) No restriction.
Given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Since we need numbers between 999 and 10000, it has only 4 digits.

Thousands place can be filled in 9 ways (excluding 0) since there is no restriction, hundreds, place, tens place and unit place can be filled in 10 ways each using all the digits.
∴ By fundamental principle of multiplication, 9 required number of 4 - digit numbers.
= 9 \(\times\) 10 \(\times\) 10 \(\times\) 10
= 9000.
(ii) No digit is repeated.

Thousands place can be filled in 9 ways (excluding 0)
Since repetition is not allowed, unit place can be filled in 9 ways tens place can be filled in 8 ways and hundreds place can be filled in 7 ways.
∴ By fundamental principle of multiplication, required number of 4 digit numbers
= 9 \(\times\) 7 \(\times\) 8 \(\times\) 9
= 4536.
(iii) At least one of the digits is repeated.
Required number of numbers = Total number of 4 digit numbers - number of 4 digit numbers when no digit is repeated
= 900 - 4536 = 4464

We know that, ⁿC₄ = 495
Therefore, \(\frac { n\times (n-1)\times (n-2)\times (n-3) }{ 4\times 3\times 2\times 1 } \) = 495
\(\Longrightarrow n \times(n-1) \times(n-2) \times(n-3)=495 \times 4 \times 3 \times 2 \times 1\)
Factoring \(495=3 \times 3 \times 5 \times 1\), and writing this product as a product of 4 consecutive numbers in the descending order we get, \(n \times(n-1) \times(n-2) \times(n-3)=12 \times 11 \times 10 \times 9\). Equating n with the maximum number, we obtain n = 12.

¹⁵C₁₃ = \(\frac { 15! }{ 2!\times 13! } =\frac { 15\times 14\times 13! }{ 2\times 1\times 13! } =\frac { 15\times 14 }{ 2\times 1 } \) = 105

When a coin is tossed, the outcomes are in two ways which are {Head, Tail}.
By the rule of product rule, the number of outcomes when 5 coins are tossed is \(2 \times 2 \times 2 \times 2 \times 2=2^{5}=32\)

Given ^(n-1)P₃ :ⁿ P₄ = 1 : 10 
⇒ \(\frac { (n-1){ P }_{ 3 } }{ n{ P }_{ 4 } } =\frac { 1 }{ 10 } \)
⇒ 10.(n-1)P₃ = 1.nP₄  \(\left[ \because npr\quad =\frac { n! }{ (n-r)! } \right] \)
⇒ \(10\times \frac { (n-1)! }{ (n-1-3)! } =\frac { n! }{ (n-4)! } \)
⇒   \(\frac { 10\times (n-1)! }{ (n-4)! } =\) \(\frac { 10\times (n-1)! }{ (n-4)! }\)⇒\(\frac { n(n-1)! }{ (n-4)! } \)
⇒10 = n
∴ n = 10.

\(\frac { 1 }{ 8! } +\frac { 1 }{ 9! } =\frac { n }{ 10! } \)
\(\frac { 1 }{ 8! } +\frac { 1 }{ 9\times 8! } =\frac { n }{ 10\times 9\times 8! } \)
Multiplying by 8! Throughout we get,
\(1+\frac { 1 }{ 9 } =\frac { n }{ 90 } \)
⇒ \(\frac { 9+1 }{ 9 } =\frac { n }{ 90 } \Rightarrow \frac { 10 }{ 9 } =\frac { n }{ 90 } \)
⇒ \(n=\frac { 10\times 90 }{ 9 } =100\)      ∴ n = 100