In how many ways, can 15 people be seated around two round tables with seating capacities of 7 and 8 people?

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

The required number of combinations, when one fruit of each kind is taken
= ⁵C₁ x ⁴C₁ x ³C₁ = 5 x 4 x 3 = 60

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

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

Total number of words = 5! = 120
combining the vowels at one place(OEA) with remaining 2 letters MG, letters can be arranged in 3! ways. Also, three vowels can be arranged in 3! ways
So, when vowels are together, then number of words = 3! x 3! = 36
there4; Required number of ways, when vowels being never together =120 - 36 = 84

Total number of letters = 8
Number of vowels = 3 and r occurs twice.
Total number of arrangements when there is no restriction = 8!/2!

When three vowels are together, regarding them as one letter, we have only 5 + 1 = 6 letters
These six letters can be arranged in 6!/2! ways, since r occurs twice.
But the three vowels can be arranged among themselves in 3! ways.
Hence number of arrangement when the three vowels are together = 6! /(2 !x 3!)

? Required number = 8!/2! - {6! / (2! x 3!)} = 18,000