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.

A number is divisible by 10 if its units digits is 0.
Therefore, 0 is fixed at the units place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places  
in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).
The 5 vacant places can be filled in 5! ways.
Hence, required number of 6-digit numbers = 5! = 120

In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times and all the other letters appear only once.
The words that will be listed before the words starting with E in a dictionary will be the words that start with A only.
Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.
Since there are 2 Is and 2 Ns in the remaining 10 letters,
Number of words starting with A = \(={10!\over 2!2!}\) \(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2\times1\times2!}\)
= 907200
Thus, the required numbers of words is 907200.

In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R.
Number of ways of selecting 2 vowels out of 3 vowels =³C₂=3
Number of ways of selecting 3 consonants out of 5 consonants = ⁵C₂=3
Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30
Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.
Hence, required number of different words = 30 × 5! = 3600

There are three servants to carry the cards. So the number of ways of sending the invitation card to the friends are 3.
So there are 3 ways of sending the invitation card to each of the five friends.
Thus required number of ways
= 3 \(\times\) 3 \(\times\)3 \(\times\)3 \(\times\) 3
= 3⁵ = 243

(2n)!=(1. 2. 3. ...n) (n+1) (n+2) (n+3) ...(2n-1) (2n) > n!. nⁿ [\(\because\) (n+r) > n for r = 1, 2, 3, ....n]
\(\therefore\) n!. nⁿ < (2n)! ...(i)
Also (n!)² = (1.2.3. ...n) (n!) \(\ge \) nⁿ. (n!) ....(ii)
Therefore, from (i) and (ii) we get  [\(\because\)  \(\underline { r } \underline { < } n\forall \underline { r } =1,\ 2,\ 3\), .....n]
(n!)² \(\le \)nⁿ.n!< (2n)! 

In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in \(=\frac { 10! }{ 3!2!2! }\) ways
Thus, required number of ways of arranging the letters of the given word
\(=\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3! }{ 3!2\times 1\times 2\times 1 } \)
= 10 \(\times\) 9 \(\times\) 8 \(\times\)7 \(\times\) 6\(\times\) 5 = 151200.

We have to find numbers between  100 and 1000, i.e. 3-digit numbers which have exactly one their digits as 7. Now, consider the following cases:
Case I When  is at unit's  place.
In this case, ten's and hundred's place can be filled by the digits 0, 1, 2, 3, 4, 5, 6, 8 and 9.
Clearly, the number of ways to fill ten's place = 9 [ by anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9]
and the number of ways to fill the hundred's place = 8
\(\therefore \) Number of such numbers  = 1 \(\times \) 9 \(\times \) 8 = 72
Case II When 7 is at ten's place.
In this case, unit's and hundred's place can be filled by the digits 0, 1, 2, 3, 4, 5, 6, 8 and 9.
Clearly, the number of ways to fill unit's place = 9 [ by anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9]
and the number of ways to fill the hundred's place = 8 [ \(\because \) hundred's place cannot be filled by zero ]
\(\therefore \) Number of such numbers = 9 \(\times \) 1 \(\times \) 8 = 72
Case III When 7 is at hundred's place.
In this case, unit's and ten's place can be filled by the digits 0, 1, 2, 3, 4, 5, 6, 8 and 9.
Clearly, the number of ways to fill unit's place = 9 [ by anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9]
and the number of ways to fill the ten's place = 9 [ by anyone of the digits 0, 1, 2, 3, 4, 5, 6, 8, 9]
\(\therefore \) Number of such numbers = 1 \(\times \) 9 \(\times \) 9 = 81
Hence, total number of required numbers
= 72 +72 + 81 =225

Lets us make the following cases and find the number of possible selection in each case.
Case I when A is chosen
In this case B most be chosen also . So we have to choose 4 person out of 6 persons, This can be done in 
\(^{ 6 }{ C }{ _{ 4 } }\) = \(^{6 }{ C }{ _{ 2 } }\) = \(\frac { 6\times 5 }{ 2\times 1 } \) = 15 ways
Case II when A is not  chosen
In this case B most be chosen also . So we have to choose 6 person out of 7 persons, This can be done in \(^{7}{ C }{ _{ 6 } }\)  = \(^{7}{ C }{ _{ 1 } }\) = 7 Ways
Hence, total number of selecting 6 persons 
=15+7=22

Let total number of perons be n
Since, total number of hand-shakes=66
\(\therefore \)  \( ^{ n }{ C }{ _{ 2 } }=\)66  \(\Rightarrow\) \(\frac { n(n-1) }{ 2 } =66\)
\(\Rightarrow\) \( { n }^{ 2 }-n-132=0\)  \(\Rightarrow\)  (n-12)(n+11)=0
\(\therefore \) n=12 [ \(\because\)  cannot be negative ]

The total number of handshakes is smae as the number of ways of selecting 2 person among 15 persons=\(^{ 15 }{ C }{ _{ 2 } }\)
\(\frac { 15\times 14 }{ 2\times 1 } =15\times 7=105\)

\(^{ 100 }{ C }{ _{ 99 } }=\frac { 100! }{ 99!(100-99)! } =\frac { 100! }{ 99!\times 1! } =\frac { 100\times 99! }{ 99!\times 1 } =100\)

We know that, \(^{n}{P}{_{r}}= r!\times^{n}{C}{{r}}\) 
\(840=r!\times 35\) 
\(\Rightarrow\)  \(r!= \frac{840}{35}\)
\(\Rightarrow\) \( r!=24=4!\) 
\(\Rightarrow\)\(r!=4\)

There are 8 distinct letters in the word TRIANGLE, out of which 3 are vowels namely A, E, I and 5 are consonants, namely T,  R, N, G, L.
(i) Since the vowels have to occur together with 5 remaining letters will be counted as 6 objects and these can be arranged in ⁶P₆ = 6 ! ways. Corresponding to each of these permutations, we have 3! = 6 permutations of the three vowels A, E and  I taken all at a time.
(ii) Clearly, by fundamental principle of multiplication, the required number of words = Number of all possible arrangements of 8 letters taken all at a time - Number of permutation in which the vowels are always together
= ⁶P₆ - 6! x 3! = 8! - 6! x 3!
= 8 x 7 x 6! - 6! x 3! = 6! ( 56-6 )
= 720 x 50 = 36000 

Let L₁, L₂, L₃, L₄, L₅ and L₆ be six letters. Since each letter can be posted in any one of the five letterboxes.
So, L₁ can be posted in 5 ways. Similarly, each of L₂, L₃, L₄, L₅ and L₆ can be posted in 5 ways.
Hence, the total number of ways in which all the six letters can be posted is 
5\(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5 = 5⁶ = 15625 ways