(i) First place can be given to anyone of the 4 children and second place can be given to anyone of the 3 remain children.
(ii) The race can be finished in 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 4! = 24 ways.

(i)Repetition of digits is allowed

The unit place can be filled in 4 ways.
Since repetition is allowed, the tens place and hundreds place can also be filled in 4 ways each.
∴ Total number one-digit numbers = 4 x 4 x 4 = 64
(ii) Repetition of digits is not allowed

The unit place can be filled in 4 ways.
Since repetition of digits is not allowed, the tens place can be filled in 3 ways.
Hundreds place can be filled in 2 ways .
∴ Total number of 3-digit numbers without repetition = 4 \(\times\) 3 \(\times\) 2 = 24

(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

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

(i) Since each question can be answered in 4 ways, the total number of ways of answering 10 questions is 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 \(\times\) 4 = 4¹⁰
(ii) Since first four question have three choices, the number of ways of answering first four questions is \(3\times 3\times 3\times 3={ 3 }^{ 4 }\)
Remaining 6 questions have 5 choices each
∴ Number of ways of answering the remaining 6 question =\(5\times 5\times 5\times 5\times 5\times 5={ 5 }^{ 6 }\)
∴ Total number of ways of answering the questions = \({ 3 }^{ 4 }\times { 5 }^{ 6 }\)
(iii) Since first four question have three choices, the number of ways of answering first four questions is \(3\times 3\times 3\times 3={ 3 }^{ 4 }\)
Remaining 6 questions have 5 choices each
∴ Number of ways of answering the remaining 6 question =\(5\times 5\times 5\times 5\times 5\times 5={ 5 }^{ 6 }\)
∴ Total number of ways of answering the questions =  \({ 3 }^{ 4 }\times { 5 }^{ 6 }\)

(i) Since each question can be answered in 4 ways, the maximum number of different answers
= \(4\times 4\times 4\times 4\times 4={ 4 }^{ 5 }\)
(ii) When each question has more than one correct answer the maximum number of different answers \(=(5 \times 3)^{5}=15^{5}\)

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

There are 11 letters in the given word of which 4 are S's,4 are I's and 2 are P's and 1 M
Hence total number of distinct words
= \(\frac { 11! }{ 4!4!2!1! } =\frac { 11! }{ 4!4!2! } \)
=\(\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 1\times 4!\times 2 } \) = 34650

LHS = ³⁵C₅ + \(\sum _{ r=0 }^{ 4\quad (39-r) }{ C_{ 4 }=^{ 40 }{ C }_{ 5 } } \)
= ³⁵C₅ + ³⁹C₄ + ³⁸C₄ + ³⁷C₄ + ³⁶C₄ + ³⁵C₄
= (³⁵C₅ + ³⁵C₄) + ³⁹C₄ + ³⁸C₄ + ³⁷C₄ + ³⁶C₄
= ³⁶C₅ + ³⁹C₄ + ³⁸C₄ + ³⁷C₄ + ³⁶C₄
= (³⁶C₅ + ³⁶C₄) + ³⁹C₄ + ³⁸C₄ + ³⁷C₄
= ³⁷C₅ + ³⁹C₄ + ³⁸C₄ + ³⁷C₄
= (³⁷C₅ + ³⁷C₄) + ³⁹C₄ + ³⁸C₄
= ³⁸C₅ + ³⁸C₃ + ³⁹C₄
= ³⁸ C₅ + ³⁹C₄ + ³⁸C₄
= ³⁹C₅  + ³⁹C₄
= ⁴⁰C₅ = RHS

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