Find the number of combinations that can be formed with 5 oranges, 4 mangoes and 3 bananas, when one fruit of each kind is taken.

Here, order is important, then the number of ways in which 12 different balls can be divided between two boys who receives 5 and 7 balls respectively, is = [12! / 5! 7! ] x 2! = 1584

Required number of straight lines
=ⁿC₂ - ^mC₂ + 1

Here, n = 10, m = 5
= ¹⁰C₂ - ⁵C₂ + 1
= 45 - 10 + 1 = 36

Hexagon has 6 sides .
? n = 6
? Required number of diagonals = ¹⁰C₂ - n
= ⁶C₂ - 6
= 6! / [2!(6 - 2)!] - 6 = 6! / (2!4!) - 6
= (6 x 5 x 4!) / (2 x 4!) - 6
= 15 - 6
= 9

Required number of triangles = ¹²C₃ = (12 x 11 x 10) / 6 = 220

Since, particular player is always chosen. It means that 11 - 1 = 10 players are selected out of the remaining 15 - 1 = 14 players.

? Required number of ways = ¹⁴C₁₀
= 14 ! / (10! x 4!)
= (14 x 13 x 12 x 11) / (4 x 3 x 2 x 1)
= 7 x 13 x 11
= 91 x 11
= 1001

Required number of ways
= 2ⁿ - 1 = 2¹⁰ - 1 = 1024 - 1 = 1023

Number of ways in which 8 persons can be selected from 15 persons = ¹⁵C₈
Now, 8 persons can be seated around a circular table in 7! ways.
Now, remaining 7 persons can be seated around a circular table in 6! ways.
? Required number of ways = ¹⁵C₈ x 7! x 6!

When O and A occupy end places. Then, the three letters (M, E, G) can be arranged themselves by 3! = 6 ways and two letters (O, A) can be arranged among themselves in 2! = 2 ways.
? Total number of ways = 6 x 2 = 12

When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways.
? Total number of ways = 4! = 24

Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place).
? Total number of ways = 3! X 2! = 12