Here ²²P_r+1: ²⁰P_r+2=11:52
\(\Rightarrow \frac { 22! }{ (21-r)! } \times \frac { (18-r)! }{ 20! } =\frac { 11 }{ 52 } \)
\(\Rightarrow \frac { 22\times 21\times 20! }{ (21-r)(20-r)(19-r)(18-r)! } \times \frac { (18-r)! }{ 20! } =\frac { 11 }{ 52 } \)
\(\Rightarrow \frac { 22\times 21 }{ (21-r)(20-r)(19-r) } =\frac { 11 }{ 52 } \)
\(\Rightarrow (21-r)(20-r)(19-r)=2\times 21\times 52\)
\(\Rightarrow (21-r)(20-r)(19-r)=14\times 13\times 12\)
\(\Rightarrow (21-r)(20-r)(19-r)\) 
= (21-7)(20-7)(19-7)
\(\Rightarrow\)  r = 7

Here \(\frac { n! }{ 5!(n-1)! } :\frac { n! }{ 7!(n-3)! } =21:1\)
\(\Rightarrow \frac { n! }{ 5!(n-1)! } \times \frac { 7!(n-3)! }{ n! } =\frac { 21 }{ 1 } \)
\(\Rightarrow \frac { 7!(n-3)! }{ 5!(n-1)! } =\frac { 21 }{ 1 } \)
\(\Rightarrow \frac { 7\times 6\times 5!(n-3)! }{ 5!(n-1)(n-2)(n-3)! } =\frac { 21 }{ 1 } \)
\(\Rightarrow \frac { 42 }{ { n }^{ 2 }-3n+2 } =\frac { 21 }{ 1 } \)
\(\Rightarrow\) 21n²-63n+42 = 42
\(\Rightarrow\) 21n(n-3) = 0
\(\Rightarrow\) n = 0 and n = 3
But when n = 0 then (n-1)! and (n-3)! have no meaning.
Thus n = 3 

\(we\quad have\quad ^{ 9 }P_{ 5 }+5.\quad ^{ 9 }P_{ 4 }=^{ 10 }P_{ r }\)
\(\Rightarrow \frac { 9! }{ 4! } +5.\frac { 9! }{ 5! } =\frac { 10! }{ (10-r)! } \Rightarrow \frac { 9! }{ 4! } +5.\frac { 9! }{ 5.4! } =\frac { 10! }{ (10-r)! } \Rightarrow 2\times \frac { 9! }{ 4! } =\frac { 10! }{ (10-r)! }\)
\( \Rightarrow \frac { 5\times 2\times 9! }{ 5\times 4! } =\frac { 10! }{ (10-r)! } \Rightarrow \frac { 10! }{ 5! } =\frac { 10! }{ (10-r)! } \Rightarrow (10-r)!=5!\)
\(\Rightarrow 10-r=5\quad \Rightarrow r=5.\)

Each signal requires the use of 2 flags.There will be as many flags as there are ways of filling in 2 vacant places  in succession by the given 5 flags of different colours.The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.
Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20

(i) There are 5 digits viz. 1, 2, 3, 4 and 5.
Since, every digit can be repeated any number of times and we have to from 3-digit numbers.So, hundred's place can be fill in 5 ways, ten's place ca be fill in 5 way and unit's place can be fill in 5 ways.

Thus, by the fundamental principle of multiplication, the number of ways in which the 3-digit number can be formed = \(5\times 5\times 5=125\) ways.
Hence, number of 3-digit numbers is 125.
(ii) When repetition of digit is not allowed.
Hundreds place can be fill in 5 ways.
Ten's place can be fill in 4 ways.
Unit's place can be ill in 3 ways.

According to FPM, total number of ways
\(=5\times 4\times 3\)
\(=60\quad ways\)
Hence, number of 3-digit numbers is 60.

n=6

Consider each married couple as one unit, then the three units can be arranged in 3! = 6 ways.
Again each couple can be arranged in 2! = 2 ways.
If the ladies sit together, then consider all ladies together as one unit. Now this unit together with 3 remaining men will be counted as 4 objects.
Ans. 48; 144 

Since the numbers between 100 and 1000 are all three digit numbers. So these three digit numbers can be formed by using the digits 0, 1, 2, 3,..., 8, 9.
Here the unit place and tens place can be filled in 10 ways each and hundreds place can be filled in 9 ways.
So total number of three digit numbers between 100 and 1000 = 9 x 10 x 10 = 900
We have to find out those numbers between 100 and 1000 which do not have 7 as a digit.
Here the unit place and tens place can be filled in 9 ways each and hundreds place can be filled in 8 ways.
So total number of three digit numbers between 100 and 1000 which do not have 7 as a digit = 8 x 9 x 9 = 648
Thus total number of three digit numbers between 100 and 1000 which having at least one of their digits as 7 = 900 - 648 = 252.

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

\(\therefore \frac { 5! }{ (5-r)! } =\frac { 6! }{ (7-r)! }\)
\( \Rightarrow \frac { 5! }{ (5-r)! } =\frac { 6\times 5! }{ (7-r)(6-r)(5-r)! }\)
\( \Rightarrow 1=\frac { 6 }{ (7-r)(6-r) } \)
\(\Rightarrow { r }^{ 2 }-13r+42=6\)
\(\Rightarrow { r }^{ 2 }-13r+36=0\)
\( \Rightarrow { r }^{ 2 }-{ 9r }-4r+36=0\)
\(\Rightarrow r(r-9)-4(r-9)=0\)
\(\Rightarrow (r-9)(r-4)=0\)
\( \Rightarrow r=9\quad or\quad r=4\)
\(Now\quad r=9\quad is\quad not\quad possible\quad because\quad r>n\)
\(Thus\quad r=4\)

From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.
Thus, required number of ways  = 
\({ }^{5} \mathrm{P}_{4}=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}\)
= 1x 2 x 3 x 4 x 5 = 120
Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4.
The number of ways in which units place is filled with digits is 2.
Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits.
Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.
Number of ways of filling the remaining places \(={ }^{4} \mathrm{P}_{3}=\frac{4 !}{(4-3) !}=\frac{4 !}{1 !}\)
= 4 × 3 × 2 × 1 = 24
Thus, by multiplication principle, the required number of even numbers is = 24 × 2 = 48