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 coats can be given to 3 men in 4 p₃ ways.5 waist coats can be given to 3 men in 5P₃ ways 6 caps can be given to 3 men in 6 P₃ ways.
∴ Total number of ways of wearing them 
= 4P3 \(\times\) 5P₃ \(\times\) 6 P₃ 
= \(\frac { 4! }{ 1! } \times \frac { 5! }{ 2! } \times \frac { 6! }{ 3! } \)
= \(4\times 3\times 2\times \frac { 5\times 4\times 3\times 2! }{ 2! } X\frac { 6\times 5\times 4\times 3! }{ 3! } \)
= 24 \(\times\) 60 \(\times\) 120 
= 1,72,800

Gold medal can be awarded to anyone of the 8 candidates in 8 ways.
Silver medal can be awarded to anyone of the remaining 7 candidates in 7 ways.
Bronze medal can be awarded to anyone of the remaining 6 candidates in 6 ways.
∴ Total numbers of ways of awarding the prize
= 8 \(\times\) 7 \(\times\) 6 = 336

Number of ways of getting a seat for 1st person = 10
Number of ways of getting a seat for 2nd person = 9
Number of ways of getting a seat for 3rd person = 8
∴ By fundamental principle of multiplication, number of ways of getting seats for 3 persons in conference hall = 10 \(\times\) 9 \(\times\) 8 = 720.

The one's place can be filled up in 5 ways using 1,2,3,4,5 and tens place can be filled up in 4 ways.
∴ Number of two digit numbers using the digits 1,2,3,4,5 is 4 x 5 = 20.

Number of ways of buying a toy car from 3 types of car = 3
Number of ways of buying a toy train from 2 types of train = 2.
∴ By fundamental principle of multiplication, number of ways of buying a toy car and a toy train = 3 \(\times\) 2 = 6.

Let P(n): = cos α + cos(α + β) + cos(α + 2β)+... + cos(α + (n - 1)β). Then,
P(1) = \(cos(\alpha )=\frac { cos(\alpha ).sin\left( \frac { \beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) } \)
Which shows P(1) is true. We now assume that P(n) is true for n=k. That is,
\(\cos \alpha +\cos (\alpha +\beta)+\cos(\alpha +2\beta)+....+\cos(\alpha+(k-1)\beta)=\cos\left(\alpha+{(k-1)\beta}\over 2 \right)\times{{\sin({k\beta\over 2})}\over{\sin({\beta\over2})}}.\)
We need to prove P(k + 1) is true, Now,
\(\underbrace { cos(\alpha )+cos(\alpha +\beta )+cos(\alpha +2\beta )+...+cos(\alpha +(k-1)\beta +cos(\alpha +k\beta ) } \)
Then P(k+1) = P(k) + cos(α + kβ)
= \(\frac { cos(\alpha +\frac { (k-1)\beta }{ 2 } )sin\left( \frac { k\beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) } +cos(\alpha +k\beta )\)
=  \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ \left( \alpha +\frac { (k-1)\beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +cos(\alpha +k\beta )sin\left( \frac { \beta }{ 2 } \right) \right] \)
= \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ cos\left( \alpha +\left( \frac { k\beta }{ 2 } \right) \right) -\frac { \beta }{ 2 } )sin\left( \alpha +\left( \frac { k\beta }{ 2 } \right) \right) +cos(\alpha +k\beta )sin\left( \frac { \beta }{ 2 } \right) \right] \)
= \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ (cos\left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) +sin\left( \alpha +\frac { k\beta }{ 2 } \right) sin\left( \frac { \beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +cos(\alpha +k\beta )sin\left( \frac { \beta }{ 2 } \right) \right] \)
= \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ cos\left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +sin\frac { \beta }{ 2 } sin\left( \alpha +\frac { k\beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +cos(\alpha +k\beta ) \right] \)
= \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ cos\left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +\frac { sin\frac { \beta }{ 2 } }{ 2 } (2sin\left( \alpha +\frac { k\beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +2cos(\alpha +k\beta ) \right] \)
= \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ cos\left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +\frac { sin\frac { \beta }{ 2 } }{ 2 } [(cos\alpha -cos(\alpha +k\beta )+2cos(\alpha +k\beta ) \right] \)
=\(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ cos\left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +\frac { sin\frac { \beta }{ 2 } }{ 2 } (cos\alpha +cos(\alpha +k\beta )) \right] \)
= \(\frac { 1 }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ cos\left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) sin\left( \frac { k\beta }{ 2 } \right) +\frac { sin\frac { \beta }{ 2 } }{ 2 } \left( 2cos \right) \left( \alpha +\frac { k\beta }{ 2 } \right) cos\left( \frac { -k\beta }{ 2 } \right) \right] \)
= \(\frac { cos\left( \alpha +\frac { k\beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) } \left[ sin\left( \frac { k\beta }{ 2 } \right) cos\left( \frac { \beta }{ 2 } \right) +sin\frac { \beta }{ 2 } cos\left( \frac { k\beta }{ 2 } \right) \right] \)
= \(\frac { cos\left( \alpha +\frac { k\beta }{ 2 } \right) sin\left( \frac { (k+1)\beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) } \)
That is, cos α + cos(α + β) + cos(α + 2β) ...+ cos(α +(k - 1)β) + cos(α + kβ)
= \(cos\left( \alpha +\frac { k\beta }{ 2 } \right) \times \frac { sin\left( \frac { (k+1)\beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) }\)
This implies that P(k + 1) is true.
The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction.
cos α + cos(α + β) + cos(α + 2β) +... cos(α+(n-1)β)
= \(cos\left( \alpha +\frac { (n-1)\beta }{ 2 } \right) \times \frac { sin\left( \frac { n\beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) } \)

Here we have to give prizes in four subjects and the process of distributing prizes can be completed by giving prizes in the four subjects.
First and second prizes can be given in mathematics in (30 \(\times\) 29) ways.
First and second prizes can be given in physics in (30 \(\times\) 29) ways.
First prize for chemistry can be given in 30 ways. First prize for English can be given in 30 ways. Hence, the number of ways to give prizes in all the four subjects = (30 \(\times\) 29) \(\times\) (30 \(\times\) 29) \(\times\) 30 \(\times\) 30.
= 6.8121 \(\times\) 10⁸

(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

\(\frac{{ }^{2 n} C_{3}}{{ }^{n} C_{3}} =\frac{11}{1} \)
\(\frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)} =\frac{11}{1} \)
\(\frac{2 n(2 n-1) 2(n-1)}{n(n-1)(n-2)} =11 \)
\(4(2 n-1) =11(n-2) \)
\(8 n-4 =11 n-22 \)
\(18 =3 n \)
\(n = 6\)

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

\(\text { Number of points of intersection }={ }^{10} \mathrm{C}_{2}\)
\(=\frac{10 \times 9}{1 \times 2}=45\)

\(\text { Number of diagonals }={ }^{n} C_{2}-n\)
\(\frac{n(n-1)}{2}-n=44 \)
\(n^{2}-n-2 n=88 \)
\(n^{2}-3 n-88=0 \)
\((n-11)(n+8)=0 \)
\(n=11(\text { or }) n=-8 \)

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

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

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

Product of r consecutive positive integers is divisible by r! (by theorem).

\({ }^{(n+5)} P_{(n+1)} \quad=\frac{11(n-1)}{2}{ }^{n+3} P_{n}\)
\(\frac{(n+5) !}{(n+5-n-1) !} =\frac{11(n-1)}{2} \times \frac{(n+3) !}{(n+3-n) !} \)
\(\frac{(n+5)(n+4)(n+3) !}{4 !} =\frac{11(n-1) \times(n+3) !}{2 \times 3 !} \)
\(\frac{(n+5)(n+4)}{4 \times 3 !} =\frac{11(n-1)}{2 \times 3 !} \)
\(n^{2}+9 n+20 =22 n-22 \)
\(n^{2}-13 n+42 =0 \)
\((n-6)(n-7) =0 \)
\(n=6 \text { or } 7 \)

4 rings can be worn in,3 fingers in 3⁴ ways

Given 9P₅ + 5.9P₄ = 10P_r
\(\Rightarrow \frac { 9! }{ 4! } +5\times \frac { 9! }{ 5! } =\frac { 10! }{ (10-r)! } \)
\(\Rightarrow \frac { 9! }{ 4! } +\frac { 9! }{ 4! } =\frac { 10! }{ (10-r)! } \quad \left[ \because \frac { 5 }{ 5! } =\frac { 5 }{ 5\times 4! } =\frac { 1 }{ 4! } \right] \)
\(\Rightarrow 2\times \frac { 9! }{ 4! } =\frac { 10! }{ (10-r)! } \)

\(\Rightarrow\) (10-r)! = 5 \(\times\)4!
\(\Rightarrow\) (10-r)! = 5!    [n(n-1)! = n!]
\(\Rightarrow\) 10-r = 5 \(\Rightarrow\) r = 5.