At least 1 question from each section is compulsory, so from the 1^st section the candidate can attempt 1 or 2 or 3 or 4 questions.
In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it.
So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.
As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 2⁴ – 1
Similarly for the 2^nd section there are 2⁵ – 1 ways in which he can attempt and for the 3^rd section there are 2⁶ – 1 ways.
The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections.
Thus, total number of ways in which he can attempt questions in that paper:
= (2⁴ – 1)(2⁵ – 1)(2⁶ – 1)
= 15 × 31 × 63
= 29295

The 5 letter word can be rearranged in 5! = 120 Ways without any of the letters repeating.
The first 24 of these words will start with A.
Then, the 25th word will start will CA _ _ _ .
The remaining 3 letters can be rearranged in 3! = 6 Ways.
i.e. 6 words exist that start with CA.
The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.
Therefore, the rank of CHASM will be 24 + 6 + 2 = 32

Case I : 2 balls of the same colour and two balls are a different colour are arranged.
Two balls of the same colour and two balls of different colours can be arranged together in which two balls of the same colour are adjacent =4!/2!x2! = 6 ways
Therefore, Total number of arrangements = 6×3 =18 ways
Case II : Two colours out of 3 can be selected in = 3C1 = 3ways
Now 2 balls of each colour can be arranged alternatively in 2 ways
Thus 4 balls can be arranged(two of each colours)
= 3×2 = 6ways
Hence total number of arrangements = 18+6 =24 ways

Case 1: 3 men and 2 women
⁸C₃*⁷C₂ = 1176
Case 2: 4 men and 1 women
⁸C₄*⁷C₁ = 490
Case 3: all 5 men
⁸C₅ = 56
Add all the cases.

All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is,
⁸C₄ = 70

In BANANA we have six letters in total but here we have some duplicate letters too so we have to deal with it and have to remove those duplicate case. B – 1 A – 3 N – 2 So total no of words possible is factorial(6) ie 6! but we must remove duplicate words: ie- (6!/(2!*3!)) which gives 60 So 60 distinguishable permutation of the letters in BANANA

The boys will be arranged in 20! ways. Now, there are a total of 21 possible places available between boys such that no 2 girls can be placed together (alternate sequence of boys and girls, starting and ending positions for girls). Therefore, the 18 girls can stand at these 21 places only. Hence, the number of ways = 20!* ²¹P₁₈

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).
We can choose any colour.
Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15

The required number of ways
(a) 1 black and 2 others = ⁴C₁.⁶C₂ = 4 × 15 = 60
(b) 2 black and 1 other = ⁴C₂.⁶C₁ = 6 × 6 = 36
(c) All the three black = ⁴C₃ = 4
Total =60 + 36 + 4 = 100

Number of ways of opting a subject other than Mathematics II = 4C2. = 4x3x2!/2!x2 = 6.

Number of ways of selection of Mathematics II = 1

Therefore, Total Number of ways = 6+1 =7.