Discussion
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- Question
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
- Correct Answer
- 108
ExplanationThe 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
- 1. 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
- 2. 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
- 3. If there are 15 dots on a circle,how many triangles can be formed?
Options- A. 455
- B. 450
- C. 469
- D. 500 Discuss
- 4. A polygon 7 sides.How many diagonals can be formed?
Options- A. 14
- B. 7
- C. 15
- D. 21 Discuss
- 5. 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
- 6. 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 Discuss
- 7. In how many ways can the letters of the word 'LEADER' be arranged ?
Options- A. 360
- B. 420
- C. 576
- D. 220 Discuss
- 8. 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 Discuss
- 9. A box contains 5 green, 4 yellow and 3 white marbles. Threemarbles are drawn at random. What is the probability thatthey are not of the same colour ?
Options- A. 40/44
- B. 44/41
- C. 41/44
- D. 40/39 Discuss
- 10. In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?
Options- A. 2800
- B. 2880
- C. 2600
- D. 3980 Discuss
Permutation and Combination problems
Search Results
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
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
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
Correct Answer: 14
Explanation:
- 7 = 14
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.
Correct Answer: 5040
Explanation:
The Word LOGARITHMS contains 10 letters.
To find how many 4 letter words we can form from that = =10x9x8x7 = 5040.
Correct Answer: 360
Explanation:
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360
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.
Correct Answer: 41/44
Explanation:
n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220
Correct Answer: 2880
Explanation:
They have to be arranged in the following way :
L T L T L T L T L
The 5 lions should be arranged in the 5 places marked ?L?.
This can be done in 5! ways.
The 4 tigers should be in the 4 places marked ?T?.
This can be done in 4! ways.
Therefore, the lions and the tigers can be arranged in 5!´ 4! ways = 2880 ways.
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