A question paper consists of two sections having respectively 3 and 5 questions. The following note is given on the paper. "It is not necessary to attempt all the question". One question from each section is compulsory. In how many ways, a candidate can select the question?

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

First , we select 13 persons out of 24 persons in ²⁴ C₁₃ ways.
Now, these 13 persons can be seated in 12! ways around a table .

So required number of ways = ²⁴ C₁₃ x 12!
= [24! / {13!(24 - 13)!}] x 12!
= [24! / {13! x 11!}] x 12!
= 24! / (13 x 11!)

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

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

Any of the 4 colour can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colours of the second stripe).
Similarly, There are 3 ways to colour each of the remaining stripes.
? The number of ways the flag can be coloured is 4(3)⁵ = (12) (3)⁴ = 12 x 81

The available digits are 0, 1, 2,...., 9. The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digit repetition not allowed). Thus, the code can be made in 9 x 9 = 81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways
Total number of ways confusion can arise
= 4 x 3 = 12
Thus, the ways in which no such confusion arise = 81-12 =69