Explanation: The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

Explanation: The given digits are 1, 3, 5, 7, 9
Sum of r digit number= n-1Pr-1
(Sum of all n digits)×(1111… r times)
N is the number of non zero digits.
Here n=5, r=4
The sum of 4 digit numbers
4P3 (1+3+5+7+9)(1111)=666600.

Explanation: First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are
54 + 54 + 54 …
54 (55times) = 54 x 55 = 2970.

Explanation:

The number of letters in the given word is four.

The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.

The word ‘LEADING’ has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

 Required number of ways = (120 x 6) = 720.

Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144

No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! – 9! × 2!
= 9!(10 – 2)
= 8 × 9!

Solution:
Taking all person of same nationality as one person, then we will have only three people.
These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.

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.

The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it.

This case can be represented as arranging ten balls and (5 – 1) four walls in the single row, which can be done in ¹⁴C₄ ways. (The balls placed between every successive pair of walls belong to one group) ¹⁴C₄ = 1001 ways.

Let n be the number of teams.
ⁿC₂ = 21
(n(n-1)/2) = 21
⇒ n(n-1) = 42 ∴
⇒ n = 7

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.