\(= { \frac { ^{ 15 }C_{ r } }{ ^{ 15 }C_{ r-1 } } }=\frac { 11 }{ 5 } \Rightarrow \frac { 15! }{ (15-r)!r! } \times \frac { (15-r+1)!\times (r-1)! }{ 15! } =\frac { 11 }{ 5 } \)
\(\Rightarrow \frac { (16-r) }{ r } =\frac { 11 }{ 5 } \Rightarrow 80-5r = 11r\)
16r = 80 ⇒ r = 5

Number of possible arrangements is same as number of permutations of 9 different things taking all at a time.
Ans.362880 

Given expression = ¹⁶C₉ - ¹⁶C₇ = ¹⁶C₇ - ¹⁶C₇
Ans. 0

We have to select 9 players out of 16 players 
11440

Given,   ⁵P_r=2\(\times \)⁶P_r-1.
\(\therefore \frac { 5! }{ (5-r)! } =2\times \frac { 6! }{ (6-r+1)! } \quad [\because \quad { ^{ n }P }_{ r }=\frac { n! }{ (n-r)! } ]\)
\(\Rightarrow \frac { 5! }{ (5-r)! } =2\times \frac { 6\times 5! }{ (7-r)! } \)
\(\Rightarrow \frac { 1 }{ (5-r)! } =\frac { 12 }{ (7-r)(6-r)(5-r)! }\)
\( \Rightarrow \frac { 1 }{ 1 } =\frac { 12 }{ (7-r)(6-r) } \)
\(\Rightarrow (7-r)(6-r)=12\)
\( \Rightarrow 42-7r-6r+{ r }^{ 2 }=12\)
\( \Rightarrow { r }^{ 3 }-13r+30=0\)
\(\Rightarrow { r }^{ 2 }-10r-3r+30=0\)
\( \Rightarrow r(r-10)-3(r-10)=0\)
\(\Rightarrow (r-3)(r-10)=0\Rightarrow r=3,10\)
\(\therefore r=3\) 

Firstly, fix the alternate position of consonant on C's position is 5! \(\times\)C1\(\times\)C2\(\times\)C3\(\times\)C4\(\times\)C5\(\times\)
Now, in any six '\(\times\)' position, 3 vowels can be arranged in ⁶P₃ ways 
Ans. 14400

We have ,\(^{n+2}{C}{_{8}}\)  :  \( ^{n-2}{P}{_{4}} \)  = 57:16

\(\Rightarrow\) \(\frac{^{n+2}{C}{_{8}} }{^{n-2}{P}{_{4}}}\) = \(\frac {57}{16}\) \(\Rightarrow \frac { \frac {(n+2)!}{8!(n+2-8)!}}{ \frac {(n-2)!}{(n-2-4)!}} = \frac {57}{16}\) 
 \(\Rightarrow\) \(\frac {(n+2)!}{8!(n-6)!} \times \frac {(n-6)!}{(n-2)!} = \frac {57}{16}\) 

 \(\Rightarrow\) \(\frac{(n+2)(n+1)n(n-1)(n-2)!}{8!} \times \frac {1}{(n-2)!} = \frac {57}{16}\) 
 \(\Rightarrow\)  (n+2)(n+1)n(n-1) = \( \frac {57}{16}\times 8!\) 
\( =\frac{19 \times 3}{16} \)  \(\times\) 8 \(\times\) 7 \(\times\)6 \(\times\)5 \(\times\)4\(\times\) 3\(\times\) 2\(\times\) 1
\(\Rightarrow\)  (n+2)(n+1)n(n-1) = 19 \(\times\)(3 \(\times\)7)\(\times\) (6\(\times\) 3 )\(\times\)(5\(\times\) 4) 
\(\Rightarrow\)  18\(\times\)19\(\times\)20\(\times\)21
On comparing both sides,we get
n-1=18 \(\Rightarrow\) n=19

We have, \(\frac { n! }{ (n-r)! } \)= \(\frac { n! }{ (n-3)! } \)  \([\because r=3,\) given\(]\)
\(=\frac { n(n-1)(n-2)(n-3)! }{ (n-3)! } \) \([\because n!=n(n-1)(n-2)(n-3)!]\)      
\(=n(n-1)(n-2)\)  

\(^{ 30 }{ C }{ _{ 0 } }=1 \quad \left[ \because ^{ \ n }{ C }{ _{ 0 } }=1 \right] \)

Here, we have six colours to colour three blocks        
So, first block can be coloured in 6 ways.
Since, each block should be coloured differently, so second block can be coloured by anyone of the remaining 5 colours.
Thus, second block can be coloured in 5 ways.
Similarly, third block can be coloured in 4 ways.
Since, each block is coloured after colouring the previous block.
\(\therefore \) By FPM, required number of designs
= 6 \(\times\) 5 \(\times\) 4 = 120 

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  

Since, the gentleman has 3 servants, so number of ways of sening the invitation card to the first sriend = 3.Similarly, for each of the remaining friends, there are 3 ways each.
Ans . Total number of ways=3⁶ = 729

In a word TRIANGLE, all letters are distrinct.
When we fix the letters T and E in the beginning and end , then rest of 6 letters can be arrangesd in 6! ways,.

Selection of three balls, consisting of atleast one black ball can be done in following ways
(i) Selecting 1 black ball and 2 non-black balls
(ii) Selecting 2 black ball and 1 non-black balls
(iii) Selecting 3 black ball and 0 non-black ball
required number od ways ³C₁ X ⁶C₂+³C₂ X⁶C₁ +³C₃
Ans. 64

Ans.36000

Required numbers will be of 1-digit, 2-digit or 3-digit.
Ans. 155