In how many ways, can 24 persons be seated around a circular table if there are is 13 seats?

There is a 7 - digit telephone number but extreme right and extreme left positions are fixed .
i. e. , 6 x x x x x 5
? Required number of ways = 8 x 7 x 6 x 5 x 4 = 6720

Married couples :
MF MF MF MF ? AB, CD, EF, CD

Possible teams :
AD CB EB GB
AF CF ED GD
AH CH EH GF

Team AD can play only with CB, CF, CH, EB, EH, GB, GF (7 teams).
Teams AD cannot play with AF, AH, ED and GD.

The same will apply with all teams, So, number of total matches = 12 x 7 = 84
But every match includes 2 teams, so the actual number of matches = 84/2 = 48

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

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

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

The required number of triangles = ⁿC₃ - ^mC₃

Here, n = 14, m = 4
= ¹⁴C₃ - ⁴C₃
= (14 x 13 x 12 x 11! ) / (3! x 11!) - 4! / (3! x 1!)
= (14 x 13 x 12)/6 - 4/1
= 14 x 26 - 4
= 364 - 4
= 360

Here, we have two sections A and B. Section A has 3 question and B has 5 question and one question from each section is compulsory according to the given condition.
? Number of ways selecting one or more than one question from section A
= 2³ - 1 = 7
Similarly, from section B = 2⁵ - 1 = 31
According to the rule of multiplication, the required number of ways in which a candidate can select the question
= 7 x 31 = 217

There are 10 station on railway line.
So, the number of different journey tickets between two station from given 10 stations from one side = ¹⁰C₂ = 10 x 9/2 = 45.

Similarly, number of different journey tickets from other side = 45
? Total number of tickets to be generated by authorities. = 45 + 45 = 90

If Mrs. X is selected among the ladies in the committee, then Mr. Y is not selected or if Mrs. X is not selected then Mr. Y can be there in the committee...
So, required number of ways
= ⁸C₃ x ⁶C₄ + ⁷C₃ x ⁷C₄
= [(8 x 7 x 6)/(3 x 2)] x [(6 x 5)/(2 x 1)] + [(7 x 6 x 5)/(3 x 2)] x [(7 x 6 x 5)/(3 x 2)]
= 840 + 1225
= 2065

A five-digit number, which is divisible by 3, is formed when sum of digits is also divisible by 3.

So, combination formed using six-digits, which are divisible by 3
= 5 + 4 + 3 + 2 + 1 = 15
= 5 + 4 + 2 + 1 + 0 = 12

So, set of number are (5, 4, 3, 2, 1) and (5, 4, 2, 1, 0).

Number formed by using 1st set = 5 x 4 x 3 x 2 x 1 = 120
Similarly, using 2nd set = 4 x 4 x 3 x 2 x 1 = 96

Hence, using 2nd set, underlined place cannot be filled by 0, otherwise it will become a four-digit number.
? Total number = 120 + 96 = 216