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.

Number of ways of getting a seat for 1st person = 10
Number of ways of getting a seat for 2nd person = 9
Number of ways of getting a seat for 3rd person = 8
∴ By fundamental principle of multiplication, number of ways of getting seats for 3 persons in conference hall = 10 \(\times\) 9 \(\times\) 8 = 720.

Since each question can be answered in 4 ways, the total number of ways of answering 6 questions is
 \(4 \times 4 \times 4 \times 4 \times 4 \times 4=4^{6}\)

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

Given ^(n-1)P₃ :ⁿ P₄ = 1 : 10 
⇒ \(\frac { (n-1){ P }_{ 3 } }{ n{ P }_{ 4 } } =\frac { 1 }{ 10 } \)
⇒ 10.(n-1)P₃ = 1.nP₄  \(\left[ \because npr\quad =\frac { n! }{ (n-r)! } \right] \)
⇒ \(10\times \frac { (n-1)! }{ (n-1-3)! } =\frac { n! }{ (n-4)! } \)
⇒   \(\frac { 10\times (n-1)! }{ (n-4)! } =\) \(\frac { 10\times (n-1)! }{ (n-4)! }\)⇒\(\frac { n(n-1)! }{ (n-4)! } \)
⇒10 = n
∴ n = 10.

Gold medal can be awarded to anyone of the 8 candidates in 8 ways.
Silver medal can be awarded to anyone of the remaining 7 candidates in 7 ways.
Bronze medal can be awarded to anyone of the remaining 6 candidates in 6 ways.
∴ Total numbers of ways of awarding the prize
= 8 \(\times\) 7 \(\times\) 6 = 336

Given factors are two a's, 3b's and 4c's
Total number of exponents = 9.
Hence, number of ways the product can be expressed without exponents
= \(\frac { 9! }{ 2!3!4! } =\frac { 9\times 8\times 7\times 6\times 5\times 4\times ! }{ 2\times 3\times 2\times 4! } \)= 1260

Here 7 players must be selected from 14 players. This can be done in 14C₇ ways.
Hence, number of different teams of players
= \(14{ C }_{ 7 }=\frac { 14! }{ 7!7! } \)
= \(\frac { 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7! }{ 7!7\times 6\times 5\times 4\times 3\times 2\times 1 } \)
= \(13\times 11\times 2\times 3\times 4\)
= 3432.

A chord is obtained by joining any two points on a circle
Number of chords drawn though 20 points is same as the number of ways of selecting 2 points out of 20 points.
This can be done in 20C₂ ways.
Hence, total number of chords is 20C₂ ​​​​​
=\(\frac { 20! }{ 2!18! } =\frac { 20\times 19\times 18! }{ 2\times 18! } =\frac { 20\times 19 }{ 2 } \)
=\(10\times 19\) = 190.

Number of subsets with no element 5C₀
Number of subsets with 1, 2, 3, 4 and 5 elements are 5C₀, 5C₁, 5C₂, 5C₃ and 5C₅ respectively.
∴ Total number of subsets
= 5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅
= \(1+5\frac { 5\times 4 }{ 2\times 1 } +\frac { 5\times 4 }{ 2\times 1 } +5+1\)
= 6 + 10 + 10 + 6
= 32