= \(\frac { 12\times 11\times 10\times 9! }{ 9!\quad 3\times 2 } =\frac { 12\times 11\times 10 }{ 6 } \)
= 2 \(\times\) 11 \(\times\) 10 = 220

The total number of signals is equal to the number of ways of filling 3 places in succession by 4 flags of different colours.
The upper place can be filled in 4 ways, following which the next place can be filled in 3 ways and the lower place can be filled in 2 ways.
Hence, by fundamental principle of multiplication, the required number of signals = 4 \(\times\) 3 \(\times\) 2 = 24.

The person can select 10 Indian food in 10 ways and 7 Chinese food in 7 ways.
∴ By fundamental principle of addition, number of ways of selecting 10 Indian or 7 Chinese food is (10 + 7) = 17 ways.

Let p(n) be the statement
\({1\over 2.5}+{1\over 5.8}+{1\over 8.11}+...+{1\over {(3n-1)(3n+2)}}={n\over 6n+4}\)
Step 1: Putting n = 1, we get
\({1\over 2.5}={1\over 6(1)+4}⇒{1\over 10}={1\over 10}\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
\(∵\ {1\over 2.5}+{1\over 5.8}+...+{1\over (3K-1)(3K+2)}={K\over 6K+4}\)
Step. 3: To show that p(K + 1) is true
ie to P.T \({1\over 2.5+}+{1\over 5.8}+...+{1\over{(3K-1)(3K+2)}}={K\over 6+4}\)
ie to P.T \({1\over 2.5}+{1\over 5.8}+...+{1\over (3K-1)(3K+2)}+{1\over (3K+2)(3K+5)}={K+1\over 6K+10}\)
LHS = \({1\over 2.5}+{1\over 5.8}+...+{1\over (3k-1)(3k+2)}+{1\over (3k+2)(3k+5)}\)
\(={k\over 6k+4}+{1\over 5.8}+...+{1\over (3k-1)(3k+2)}+{1\over (3K+2)(3k+5)}\)
\(={k\over 6k+4}+{1\over (3k+2)(3k+5)}\) [using (1)]
\(={k\over 2(3k+2)}+{1\over (3k+2)(3k+5)}\)
\(={1\over 3k+2}\left[{k\over 2}+{1\over 3k+5}\right]\)
\(={1\over 3k+2}\left[k(3k+5)+2\over 2(3k+5)\right]={1\over 3k+2}\left[3k^2+5k+2\over 2(3k+5)\right]\)
\(={1\over 2(3k+5)(3k+5)}(k+1)(3k+2)\)
\(={k+1\over 2(3k+5)}={k+1\over 6k+10}=RHS\)

∴ p (k + 1) is true.
Hence, by the principle 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
= 9000 - 4536 = 4464

Number of ways of selecting 8 people from 12 in ¹²C₈ ways.
Let A and B both attend the party
Out of 10 remaining people 8 can attend  in ¹⁰C₆ ways.
.'. Number of ways in which two of them do not attend together = ¹²C₈ - ¹⁰C₆

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).

First in Maths = 30 ways
Second in Maths - 29 ways
Similarly for other subjects
30 \(\times\) 29 \(\times\) 30 \(\times\) 29 \(\times\) 30 \(\times\)30 = 30⁴ \(\times\)29²

Total numbers = 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24,
Sum of all integers in tenth place
= 6 ( 2 + 4 + 5 + 7) = 108

Let there be n sides of the polygon. We know that the number of diagonals of n sided polygon is \(\frac { n(n-3) }{ 2 } \)
⇒ Given \(\frac { n(n-3) }{ 2 } =90\)
⇒ n²-2n = 180
⇒ n²-3n-180 = 0
⇒ (n-15) (n+12) = 0
⇒  n = 15 or n = -12
⇒ There are 15 sides for the polygon which has 90 diagonals.

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

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