A five-digit number divisible by 3 is to be formed using digits 0, 1, 2, 3, 4 and 5 without repetition, the total number of ways this can be done, is

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

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

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

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!)

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

The digit in the unit's place should be greater than that in the tens' place.
Hence, if digit 5 occupies the unit place, then remaining four digits need not to follow any order,hence required number = 4!
However, if digit 4 occupies the unit place then 5 cannot occupy the ten;s position. Hence, digit at the ten's place and it will be filled by the digit 1, 2 or 3. This can happen in 3 ways. The remaining 3 digit can be filled in the remaining three place in 3! ways.
Hence, in all, we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit's place and it will be either 1 or 2. this can happen in 2 ways. Hence, we will have (2 x 3! ) number ending in 3 . Similarly, we can find that there will be 3! numbers ending in 2 and no number with 1. Hence, total number of numbers
= 4! + (3) x 3! + (2 x 3!) + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60

Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4, and 5) with either blue or red balls 2⁵ = 32.
Two adjacent boxes with blue can be obtained in 4 ways, i.e., (12), (23), (34) and (45).
Three adjacent boxes with blue can be obtained in 3 ways, i.e., (123), (234)and (345). Four boxes with blue can be obtained in 2 ways, i.e., (1234) and (2345). And five boxes with blue can be got in 1 way. Hence, the number of ways of filling the boxes such that adjacent boxes have blue
= (4 + 3 + 2 + 1) = 10.
Hence, the number or ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22.

27 is in the middle, so is the median and its the mean too.