An intelligence agency forms a code of two distinct digits selected from 0, 1, 2,....., 9 such that the first-digit of code is non-zero. The code, handwritten on a slip, can however potentially create confusion when read upside down, for examples the code 91 may appear as 16. how many codes are there for which no such confusion can arise?

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

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

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 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.

4th observation i.e., 8 is the median.

Anuj reached the place of meeting at 8:15 h, he reached 30 min earlier than the man who was 40 min late, Hence, the scheduled time of the meeting was 8:05 h.