99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1
\(\therefore(99)^{5}=(100-1)^{5} \)
\(={ }^{5} \mathrm{C}_{0}(100)^{5}-{ }^{5} \mathrm{C}_{1}(100)^{4}(1)+{ }^{5} \mathrm{C}_{2}(100)^{3}(1)^{2}-{ }^{5} \mathrm{C}_{3}(100)^{2}(1)^{3} +{ }^{5} \mathrm{C}_{4}(100)(1)^{4}-{ }^{5} \mathrm{C}_{5}(1)^{5} \)
\(=(100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}+5(100)-1 \)
\(= 10000000000-500000000+10000000-100000+500-1\)
\(= 10010000500-500100001 \)
\(=9509900499\)

Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different.
Therefore, the required number of arrangements 
\(=\frac{9 !}{4 ! 2 !}=\frac{5 \times 6 \times 7 \times 8 \times 9}{2}=7560\) 

Here \({ a }_{ n }=\frac { 2n-3 }{ 6 } \)
Putting n = 1, 2, 3, 4 and 5, we have
\({ a }_{ 1 }=\frac { 2\times 1-3 }{ 6 } =\frac { 2-3 }{ 6 } =\frac { -1 }{ 6 } \)
\({ a }_{ 2 }=\frac { 2\times 2-3 }{ 6 } =\frac { 4-3 }{ 6 } =\frac { 1 }{ 6 } \)
\({ a }_{ 3 }=\frac { 2\times 3-3 }{ 6 } =\frac { 6-3 }{ 6 } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\({ a }_{ 4 }=\frac { 2\times 4-3 }{ 6 } =\frac { 8-3 }{ 6 } =\frac { 5 }{ 6 } \)
\({ a }_{ 5 }=\frac { 2\times 5-3 }{ 6 } =\frac { 10-3 }{ 6 } =\frac { 7 }{ 6 } \)
Thus, first five terms of sequence are: \(-\frac { 1 }{ 6 } ,\frac { 1 }{ 6 } ,\frac { 1 }{ 2 } ,\frac { 5 }{ 6 } and\frac { 7 }{ 6 } \) 

By using Binomial Theorem, the expression (1– 2x)⁵ can be expanded as,
\((1-2x)^{ 5 }=\left[ 1+\left( -2x \right) \right] ^{ 5 }\)
\(=^{ 5 }{ C }_{ o }+^{ 5 }{ C }_{ 1 }(-2x)+^{ 5 }{ C }_{ 2 }(-2x)^{ 2 }+^{ 5 }{ C }_{ 3 }(-2x)^{ 3 }+^{ 5 }{ C }_{ 4 }(-2x)^{ 4 }+^{ 5 }{ C }_{ 5 }(-2x)^{ 5 }\)
\(=1+5(-2x)+10({ 4x }^{ 2 })+10({ -8x }^{ 3 })+5({ 16x }^{ 4 })+1(-{ 32x }^{ 5 })\)
\(=1-10x+{ 40x }^{ 2 }-{ 80x }^{ 3 }+{ 80x }^{ 4 }-{ 32x }^{ 5 }\)
which is the required expansion.

In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in which there is exactly one ace. Then, one ace can be selected in  ⁴C₁ ways and the remaining 4 cards can be selected out of the 48 cards in 48C4ways. Thus, by multiplication principle, required number of 5 card combinations
 ⁴C₁ x ⁴⁸C₄ \(=\frac{48 !}{4 ! 44 !} \times \frac{4 !}{1 ! 3 !}\)
\(={4\times3!\over 3!}\times{48\times47\times46\times45\times44!\over 4\times3\times2\times44!}\)
= 778320

Proof is very similar to that of Theorem 1 and is left for the reader to arrive at. Here, we are solving some of the problems of the pervious Section using the formula for nPr to illustrate its usefulness. In Example 1, the required number of words = ⁴P₄ = 4! = 24. Here repetition is not allowed. If repetition is allowed, the required number of words would be 4⁴ = 256. The number of 3-letter words which can be formed by the letters of the word NUMBER \(={ }^6 \mathrm{P}_3=\frac{6 !}{3 !}=4 \times 5 \times 6=120\). Here, in this case also, the repetition is not allowed. If the repetition is allowed, the required number of words would be \(6^3=216\). The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly \({ }^{12} \mathrm{P}_2=\frac{12 !}{10 !}=11 \times 12=132\)

Total letters in the word PERMUTATIONS = 12.
Here T = 2.
(i) Now first letter is P and last letter is S, which are fixed.
So the remaining 10 letters are to be arranged between P and S
∴ Number of permutations = \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
∴ Number of permutations of all vowels together = ⁵p₅
\(={5!\over 0!}=5 \times 4 \times 3 \times 2 \times 1 = 120\)
Now consider the 5vowels together as one letter. Sothe number of letters in the word when all vowels are together = 8.
∴ Number of permutations  \(={8!\over 2!}\)
\(={8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=20160
Hence the total number of permutations
= 120 x 20160 = 2419200
(iii) Here P and S are on 1st and 6th places
P and S are on 2nd and 7th places
P and S are on 3rd and 8th places P and S are on 4th and 9th places
P and S are on 5th and 10th places P and S are on 6th and 11th places
P and S are on 7th and 12th places
Now we see that P and S can be put in 7 ways and also P and S can interchange their positions.
∴ Number of permutations = 2 x 7 = 14
Now the remaining 10 places can be filled with remaining 10 letters.
∴ Number of permutations =  \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
Thus total number of permutations = 14 x 1814400= 25401600

We have (3x² - 2ax + 3a²)³
=[(3x²- 2ax) + 3a²)]³
=³C₀(3x²- 2ax)³ + ³C₁(3x² - 2ax)² (3a²)+³C₂(3x²- 2ax) (3a²)²+ ³C₃(3a²)³
= (3x² - 2ax)³ + 3 x 3a² (3x² - 2ax)² + 3x 9a⁴ (3x² - 2ax) + 27a⁶
= (27x⁶ - 8a³x³ - 54ax⁵ + 36a²x⁴) + 9a² (9x⁴+ 4a²x² - 12ax³) + 27a⁴ (3x² - 2ax) + 27a⁶
= 27x⁶ - 108a³x³- 54ax⁵ + 36a²x⁴ + 81a²x⁴+ 36a⁴x² - 108a³x³ + 81a⁴x² - 54a⁵x + 27a⁶
= 27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³
+ 117a⁴x² - 54a⁵x + 27 a⁶

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 [1 + 11 + 111 + 1111 +.... upto n terms]
= \(\frac { 8 }{ 9 } \) [9 + 99 + 999 + 9999 + ... upto n terms]
= \(\frac { 8 }{ 9 } \) [(10 - 1)+(10² - 1) + (10³ - 1)] +....upto n terms
= \(\frac { 8 }{ 9 } \) [(10 + 10² +10³ + ...n terms) - (1+1+...n terms)]
= \(\frac { 8 }{ 9 } \left[ \frac { 10.\left( { 10 }^{ n }-1 \right) }{ \left( 10-1 \right) } -n \right] \)
= \(\frac { 8 }{ 9 } \left[ \frac { 10 }{ 9 } \left( { 10 }^{ n }-1 \right) -n \right] \)
= \(\frac { 80 }{ 81 } \left[ { 10 }^{ n }-1 \right] -\frac { 8 }{ 9 } n\)

Given, \(\frac { { a }^{ n+1 }+{ b }^{ n+1 } }{ { a }^{ n }+{ b }^{ n } } =\frac { { a }^{ \frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } } }{ 1 } \)
\(\Rightarrow { a }^{ n+1\quad }+{ b }^{ n+1 }={ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }+{ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } }\)
\(\Rightarrow ({ a }^{ n+1 }-{ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } })+({ b }^{ n+1 }-{ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } })=0\)
\(\Rightarrow ({ a }^{ n+\frac { 1 }{ 2 } }-{ b }^{ n+\frac { 1 }{ 2 } })({ a }^{ \frac { 1 }{ 2 } }-{ b }^{ \frac { 1 }{ 2 } })=0\)
\(\Rightarrow { a }^{ n+\frac { 1 }{ 2 } }- { b }^{ n+\frac { 1 }{ 2 } }=0\)

In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4 consonants, namely, N, V, L and T.
The number of ways of selecting 3 vowels out of 4 = ⁴C₃ = 4.
The number of ways of selecting 2 consonants out of 4 = ⁴C₂ = 6. Therefore, the number of combinations of 3 vowels and 2 consonants is 4 × 6 = 24.
Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is 24 × 5 ! = 2880

Let \(\frac {a}{r}\) a, ar be the first three terms of the G.P. Then
\(\frac{a}{r}+a r+a=\frac{13}{12} \)
and  \(\left(\frac{a}{r}\right)(a)(a r)=-1\)
From (2), we get a³ = – 1, i.e., a = – 1 (considering only real roots)
Substituting a = –1 in (1), we have
\(-\frac{1}{r}-1-r=\frac{13}{12} \text { or } 12 r^{2}+25 r+12=0\)
This is a quadratic in r, solving, we get \(r=-\frac{3}{4} \text { or }-\frac{4}{3}\)
Thus, the three terms of G.P. are : \(\frac{4}{3},-1, \frac{3}{4} \text { for } r=\frac{-3}{4} \text { and } \frac{3}{4},-1, \frac{4}{3} \text { for } r=\frac{-4}{3}\)

We have, \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}=\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}\)
\(={ }^{4} C_{0}^{\circ}\left(1+\frac{x}{2}\right)^{4}-{ }^{4} C_{1}\left(1+\frac{x}{2}\right)^{3}\left(\frac{2}{x}\right)+{ }^{4} C_{2}\left(1+\frac{x}{2}\right)^{2} \)
\(\left(\frac{2}{x}\right)^{2}-{ }^{4} C_{3}\left(1+\frac{x}{2}\right)\left(\frac{2}{x}\right)^{3}+{ }^{4} C_{4}\left(\frac{2}{x}\right)^{4} \)
\(=\left(1+\frac{x}{2}\right)^{4}-4\left(1+\frac{x}{2}\right)^{3} \frac{2}{x}+\frac{4 \times 3}{2}\left(1+\frac{x}{2}\right)^{2} \frac{4}{x^{2}} \)
\(-4\left(1+\frac{x}{2}\right) \times \frac{8}{x^{3}}+\frac{16}{x^{4}} \)
\(=\left(1+\frac{x}{2}\right)^{4}-\frac{8}{x}\left(1+\frac{x}{2}\right)^{3}+\frac{24}{x^{2}}\left(1+\frac{x}{2}\right)^{2}\)
\(-\frac{32}{x^{3}}\left(1+\frac{x}{2}\right)+\frac{16}{x^{4}}\)
Now, on expending \(\left(1+\frac{x}{2}\right)^{4},\left(1+\frac{x}{2}\right)^{3} \cdot\left(1+\frac{x}{2}\right)^{2}\) 
We get, \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}=\left(1+4 \cdot \frac{x}{2}+6 \cdot \frac{x^{2}}{4}+4 \cdot \frac{x^{3}}{8}+\frac{x^{4}}{16}\right) \)
\(-8 \cdot \frac{1}{x}\left(1+3 \cdot \frac{x}{2}+3 \cdot \frac{x^{2}}{4}+\frac{x^{3}}{8}\right) \)
\(+24 \cdot \frac{1}{x^{2}}\left(1+x+\frac{x^{2}}{4}\right)-32 \times \frac{1}{x^{3}}\left(1+\frac{x}{2}\right)+\frac{16}{x^{4}} \)
\(=\left(1+2 x+\frac{3 x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}\right)-\left(\frac{8}{x}+12+6 x+x^{2}\right) \)
\(+\left(\frac{24}{x^{2}}+\frac{24}{x}+6\right)-\left(\frac{32}{x^{3}}+\frac{16}{x^{2}}\right)+\frac{16}{x^{4}} \)
\(=\frac{x^{4}}{16}+\frac{x^{3}}{2}+x^{2}\left(\frac{3}{2}-1\right)+x(2-6)+(1-12+6) \)
\(+(24-8) \frac{1}{x}+(24-16) \frac{1}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}} \)
\(=\frac{x^{4}}{16}+\frac{x^{3}}{2}+\frac{x^{2}}{2}-4 x-5+\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}\)

Total cost of the tractor = Rs12000
Cash amount paid = Rs 6000
Balance amount = Rs12000 - Rs 6000 = Rs 6000
Interest of 1st instalment =Rs\(\frac { 6000\times 12\times 1 }{ 100 } \)=Rs 720
Amount of 1st instalment =Rs500+Rs720=Rs1220
Interest of second instalment =Rs \(\frac { 5500\times 12\times 1 }{ 100 } \)=Rs 660
Amount of second instalment =Rs 500+Rs 600=Rs1160
Interest of the 3rd instalment \(=\frac { 5000\times 12\times 1 }{ 100 } \)=Rs 600
Amount of third instalment = 500 + 600 = Rs1100
The sequence of instalment is 1220, 1160, 1100 ...
Here a = 1220 d = 1160 - 1220 = - 60 and n = 12
∴ S_n=\(\frac { n }{ 2 } \)[2a+(n-1)d]
∴ S₁₂=\(\frac { 12 }{ 2 } \)[2 x 1220+(12-1) x -60]
=6[2400-660]=Rs 10680
Thus total cost of tractor =10680+6000=Rs 16680