Permutation and Combination problems
- 1. There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairytales, novels and plays ?
Options- A. 12400
- B. 17820
- C. 17280
- D. 12460 Discuss
Correct Answer: 17280
Explanation:
There are 4 books on fairy tales and they have to be put together. They can be arranged in 4! ways.
Similarly, there are 5 novels.They can be arranged in 5! ways.
And there are 3 plays.They can be arranged in 3! ways.
So, by the counting principle all of them together can be arranged in 4!´5!´3! ways = 17280
- 2. How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
Options- A. 1000
- B. 100
- C. 500
- D. 999 Discuss
Correct Answer: 100
Explanation:
1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ?n?.
Therefore, using ?n? alphabets we can form n * n * n = distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This different initials = 1 million
i.e. (1 million = )
=> n = = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
- 3. A polygon 7 sides.How many diagonals can be formed?
Options- A. 14
- B. 7
- C. 15
- D. 21 Discuss
Correct Answer: 14
Explanation:
- 7 = 14
- 4. If there are 15 dots on a circle,how many triangles can be formed?
Options- A. 455
- B. 450
- C. 469
- D. 500 Discuss
Correct Answer: 455
Explanation:
There are 15 dots in total,and to make a triangle we need to select any three of those dots.
So, = 455
- 5. When six fairs coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads ?
Options- A. 15
- B. 42
- C. 16
- D. 40 Discuss
Correct Answer: 42
Explanation:
The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.
i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.
The number of outcomes in which 0 coins turn heads is =1
The number of outcomes in which 1 coin turns head is = =6
The number of outcomes in which 2 coins turn heads is =15
The number of outcomes in which 3 coins turn heads is =20
Therefore, total number of outcomes =1+6+15+20= 42 outcomes
- 6. There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?
Options- A. 70
- B. 40
- C. 72
- D. 80 Discuss
Correct Answer: 70
Explanation:
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: 8 = 70
- 7. If the letters of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word VERMA is :
Options- A. 108
- B. 117
- C. 810
- D. 180 Also asked in: CAT, Bank Exams, AIEEE, Bank PO, Bank Clerk
Correct Answer: 108
Explanation:
The number of words beign with A is 4!
The number of words beign with E is 4!
The number of words beign with M is 4!
The number of words beign with R is 4!
Number of words beign with VA is 3!
Words beign with VE are VEAMR
VEARM
VEMAR
VEMRA
VERAM
VERMA
Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108
- 8. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed ?
Options- A. 5040
- B. 2525
- C. 2052
- D. 4521 Also asked in: GRE, GATE, CAT, Bank Exams, AIEEE, Bank PO, Bank Clerk
Correct Answer: 5040
Explanation:
The Word LOGARITHMS contains 10 letters.
To find how many 4 letter words we can form from that = =10x9x8x7 = 5040.
- 9. In how many ways can the letters of the word 'LEADER' be arranged ?
Options- A. 360
- B. 420
- C. 576
- D. 220 Also asked in: AIEEE, Bank Exams, CAT, Bank Clerk
Correct Answer: 360
Explanation:
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360
- 10. In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?
Options- A. 1,51,200 ways.
- B. 5,04,020 ways
- C. 72,000 ways
- D. None of the above Also asked in: AIEEE, Bank Exams, CAT, GATE, Analyst, Bank Clerk, Bank PO
Correct Answer: 1,51,200 ways.
Explanation:
In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E
Total = 13 letters
But last letter must be N
Hence, available places = 12
In that odd places = 1, 3, 5, 7, 9, 11
Owvels = 4
This can be done in 6P4 ways
Remaining 7 letters can be arranged in 7!/3! x 2! ways
Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.
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