Here ⁿP_r  = ⁿP_r+1
\(\Rightarrow \frac { n! }{ (n-r)! } =\frac { n! }{ (n-r-1)! } \)
\(\Rightarrow \frac { 1 }{ (n-r)(n-r-1)! } =\frac { 1 }{ (n-r-1)! } \)
\(\Rightarrow \frac { 1 }{ n-r } =1 \Rightarrow n-r=1\) ....(i)
Also ⁿC_r = ⁿC_r-1
\(\Rightarrow \frac { n! }{ (n-r)!r! } =\frac { n! }{ (n-r+1)!(r-1)! } \)
\(\Rightarrow \frac { 1 }{ (n-r)!(r-1)! } =\frac { 1 }{ (n-1+1)(n-r)!(r-1)! } \)
\(\Rightarrow \frac { 1 }{ r } =\frac { 1 }{ n-r+1 } \)
\(\Rightarrow \)  n-r+1=r \(\Rightarrow \) n - 2r = -1  ....(ii)
From (i) and (ii), we have
n = 3 and r = 2.

40

1

\(\frac { 13 }{ 8064 } \)

30030

Firstly, fix the odd position of vowels (A, I, E) 1, 2, 3, 4, 5, 6, 7 i.e.
⁴P₃ ways and sent of the 5 places can be arranged in 5! ways.
Ans. 576

We have, \(\frac { 8! }{ 4! } \) =\(\frac { 8\times 7\times 6\times 5\times 4! }{ 4! } \) \([\because n!=n(n-1)(n-2)....3.2.1]\)
\(=8\times 7\times 6\times 5=1680\)
Also, \(2!=2\times 1=2\neq 1680\)
\(\therefore \)  \(\frac { 8! }{ 4! } \neq 2!\) 

Let us take persons of same nationally as one unit. Then, there are 3 units which can be arranged in 3! = 6 ways. Now, in each of arrangements, 8 Indians can be arranged among themselves in 8! ways, 4 Americans can be arranged among themselves in 4! ways and 4 Englishmen can be arranged among themselves in 4! ways.
Hence, by the fundamental principle of multiplication, required a number of ways = 3! \(\times \)8! \(\times \)4!  \(\times \)4!.

In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.
Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be ²P₂=2!
Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.
Hence, by multiplication principle, required number of words = 2! × 5! × 3!
= 1440

Here, we need to perform two operations:
(i)  Selecting a door to enter.
(ii)  Selecting a door to come out.
Clearly, the man can enter the room through anyone of the seven doors. So, there are seven ways of entering into the room. Note that the man can come out through anyone of the remaining six doors. So, he can come out through a different door in 6 ways. Hence, by fundamental principle of counting, required number of ways = 7 \(\times\) 6 = 42  

There will be as many ways as there are ways of filling 2 vacant places  in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.

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

(i) Each person can leave the cabin at anyone of the seven floors. So, each person can leave the cabin in 7 ways.
Ans. 7⁵
(ii) First person can leave the cabn at anyone of the seven floors. Second person can leave the cabin at anyone of the remaining 6 floors. Similarly, we can calculate for other.
Ans. 2520

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

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