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)
= \(cos\left( \alpha +\frac { (n-1)\beta }{ 2 } \right) \times \frac { sin\left( \frac { n\beta }{ 2 } \right) }{ sin\left( \frac { \beta }{ 2 } \right) } \)

(i) There are 8 letters in the word "EQUATION" which includes 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). Considering 5 vowels as one letter, we have 4 letters which can be arranged in ⁴P₄= 4! ways. But corresponding each of these arrangements, the vowels E, U, A, I, O can be put in ⁵P₅ = 5! ways.
Hence, by the rule of product required number of words is 4! \(\times\) 5! = 24 \(\times\) 120 = 2880.
(ii) The total number of strings formed by using all the eight letters of the word "EQUATION" is ⁸P₈ = 8! = 40320.
So, the total number of strings in which vowels are never together is the same as the difference between the total number of strings and the number of strings in which vowels are together is 40320 - 2880 = 37440.

\(\frac { (2n)! }{ n! } =\frac { 1.2.3.4...(2n-2).(2n-1)2n }{ n! } \)
= \(\frac { (1.3.5...(2n-1))(2.4.6...(2n-2).2n) }{ n! } \) (Grouping the odd and even numbers separately)
= \(\frac { (1.3.5...(2n-1))\times 2^{ n }\times (1.2.3...(n-1).n) }{ n! } \) (taking out the 2's)
= \(\frac { (1.3.5...(2n-1)\times 2^{ n }\times n! }{ n! } \)
= 2ⁿ(1.3.5...(2n - 1))

Three balls must be drawn with atleast one black ball.
The possible choices are as follows

∴ Required number of ways of drawing 3 balls?
= \(2{ C }_{ 2 }\times { 3C }_{ 1 }\times { 4C }_{ 0 }+2C_{ 0 }\times { 3C }_{ 1 }\times { 4C }_{ 2 }+2C_{ 1 }\times { 3C }_{ 1 }\times { 4C }_{ 1 }\)\(\times \) \({ 2C }_{ 1 }\times 3C_{ 2 }\times { 4C }_{ 0 }+{ 2C }_{ 0 }\times { 3C }_{ 2 }\times { 4C }_{ 1 }+{ 2C }_{ 0 }\times { 3C }_{ 3 }\times { 4C }_{ 0 }\)
= 1\(\times \)3\(\times \)1+1\(\times \)3\(\times \)6+2\(\times \)3\(\times \)4+2\(\times \)3\(\times \)1+1\(\times \)3\(\times \)4+1\(\times \)1
= 3 + 18 + 24 + 6 + 12 + 1 = 64

The different possible arrangements are given in the following table

Hence, die required number of ways
= (4C₃) (4C₃) + (4C₂) (3C₁) (3C₁) (4C₂) + (4C₁) (3C₂) (3C₂) (4C₁) + (3C₃) (3C₃)
= (4) (4) + (6) (3) (3) (6) + 4 (3) (3) (4) + 1(1)
= 16 + 324 + 144 + 1
= 485

(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) Assume that no three points out of 11 points are collinear. Then we can draw unique straight through any arbitrary pair of points out of the 11 given points. This is a combination of 2 objects taken at a time from a total of 11 and can be done in 11C₂ = 55 ways.
However, it is given that 4 points are collinear. Had they not been collinear the number of unique lines that could have been drawn through them is a combination of 2 objects taken at a time from a total of 5 and can be done in 4C₂ = 6 ways. Since they are collinear we get only one line out of these 4 points instead of 6.
So, the total number of straight lines that can be drawn through 13 points on a plane with 5 of the points being collinear is 55 - 6 + 1 = 50.
(ii) To form a triangle we need 3 points. The following are the choices.
a) If we take one point from 4 collinear points and 2 from remaining 7. So this case will give \({ 4C }_{ 1 }\times { 7C }_{ 2 }\) points = \(4\times 21=84\)
b) If we two points from 4 collinear points and 1 from remaining 7. So this will give \({ 4C }_{ 2 }\times { 7C }_{ 1 }=6\times 7=42\)
c)If we take all the three points from 7 non collinear points. Which will give 7C₃ = 35
∴ Total number of triangles are 84 + 42 + 35 = 161.

Examination has 11 letters, and in which 'A' 'I' and 'N', all occur twice 11C4' would have been fine if all letters were distinct. So we have E, X, M, T, 0, (AA), (II), (NN). 8 distinct letters.
\( { }^{8} \mathrm{C}_{4} \times 4 ! =\frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4} \times 4 \times 3 \times 2 \times 1 \)
\( =1680 \)
\({ }^{3} C_{1} \times{ }^{7} C_{2} \times \frac{4 !}{2 !} =3 \times \frac{7 \times 6}{1 \times 2} \times \frac{4 \times 3 \times 2 \times 1}{2 \times 1}=756 \)
\({ }^{3} C_{2} \times \frac{4 !}{2 ! 2 !} =\frac{3 \times 2}{1 \times 2} \times \frac{4 \times 3 \times 2 \times 1}{2 \times 2}=18 \)
Total no. of ways = 2454 

The number of 4-digit numbers that can be formed using the 5 digits is ⁵P₄ = 120
Let us find the sum of the digits in the unit place.
 \(=5 \times 4 \times 3 \times 2=20 \times 6\)
= 120 numbers
Each of the five given numbers will be repeated 24 times. Hence, sum of digits appearing in any place
= 24 (1 + 2 + 3 + 4 + 5)
24 \(\times\) 15 = 360
Sum of all 4 digits = 360 (1000 +100 + 10 + 1)
360 \(\times\) 1111 = 399960

In the word THING, there are 5 letters
The lexicographic order of the word is G, H, I,N, T
Number of words starting with G = 4! = 24
Number of words starting with H =  4! = 24
Number of words starting with I 4! = 24
Number of words starting with NG 3! = 6
Number of words starting with NGH 2! = 2
Number of words starting with NGHI = 1!
Number of words starting with NGHIT = 1!
85^th string NGHIT