Permutation and Combination problems
- 1. Find the number of combinations that can be formed with 5 oranges, 4 mangoes and 3 bananas, when one fruit of each kind is taken.
Options- A. 60
- B. 120
- C. 110
- D. 75 Discuss
Correct Answer: 60
Explanation:
The required number of combinations, when one fruit of each kind is taken
= 5C1 x 4C1 x 3C1 = 5 x 4 x 3 = 60
- 2. There are 10 questions in a question paper. In how many ways, a student can solve there questions, if he solves one or more question?
Options- A. 1024
- B. 1025
- C. 1023
- D. 1000 Discuss
Correct Answer: 1023
Explanation:
Required number of ways
= 2n - 1 = 210 - 1 = 1024 - 1 = 1023
- 3. In how many ways, can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
Options- A. 15!/(8!)
- B. 7!/88!
- C. 15C8 x 6! x 7!
- D. 15C8 x 8! Discuss
Correct Answer: 15C8 x 6! x 7!
Explanation:
Number of ways in which 8 persons can be selected from 15 persons = 15C8
Now, 8 persons can be seated around a circular table in 7! ways.
Now, remaining 7 persons can be seated around a circular table in 6! ways.
? Required number of ways = 15C8 x 7! x 6!
- 4. O and A occupying end places.
Options- A. 12
- B. 14
- C. 20
- D. 18 Discuss
Correct Answer: 12
Explanation:
When O and A occupy end places. Then, the three letters (M, E, G) can be arranged themselves by 3! = 6 ways and two letters (O, A) can be arranged among themselves in 2! = 2 ways.
? Total number of ways = 6 x 2 = 12
- 5. E being always in the middle.
Options- A. 18 ways
- B. 24 ways
- C. 48 ways
- D. 20 ways Discuss
Correct Answer: 24 ways
Explanation:
When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways.
? Total number of ways = 4! = 24
- 6. Vowels occupying odd places .
Options- A. 12 ways
- B. 16 ways
- C. 6 ways
- D. 20 ways Discuss
Correct Answer: 12 ways
Explanation:
Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place).
? Total number of ways = 3! X 2! = 12
- 7. Vowels being never together.
Options- A. 36 ways
- B. 84 ways
- C. 120 ways
- D. 10 ways Discuss
Correct Answer: 84 ways
Explanation:
Total number of words = 5! = 120
combining the vowels at one place(OEA) with remaining 2 letters MG, letters can be arranged in 3! ways. Also, three vowels can be arranged in 3! ways
So, when vowels are together, then number of words = 3! x 3! = 36
there4; Required number of ways, when vowels being never together =120 - 36 = 84
- 8. In how many ways can the letters of the word 'Director' be arranged so that the three vowels are never together?
Options- A. 1800
- B. 18000
- C. 16000
- D. 1600 Discuss
Correct Answer: 18000
Explanation:
Total number of letters = 8
Number of vowels = 3 and r occurs twice.
Total number of arrangements when there is no restriction = 8!/2!
When three vowels are together, regarding them as one letter, we have only 5 + 1 = 6 letters
These six letters can be arranged in 6!/2! ways, since r occurs twice.
But the three vowels can be arranged among themselves in 3! ways.
Hence number of arrangement when the three vowels are together = 6! /(2 !x 3!)
? Required number = 8!/2! - {6! / (2! x 3!)} = 18,000
- 9. A mixed doubles tennis game is to be played between two teams ( each team consists of one male and one female). There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?
Options- A. 12
- B. 48
- C. 36
- D. 42 Discuss
Correct Answer: 48
Explanation:
Married couples :
MF MF MF MF ? AB, CD, EF, CD
Possible teams :
AD CB EB GB
AF CF ED GD
AH CH EH GF
Team AD can play only with CB, CF, CH, EB, EH, GB, GF (7 teams).
Teams AD cannot play with AF, AH, ED and GD.
The same will apply with all teams, So, number of total matches = 12 x 7 = 84
But every match includes 2 teams, so the actual number of matches = 84/2 = 48
- 10. There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone number are possible?
Options- A. 120
- B. 100000
- C. 6720
- D. 30240 Discuss
Correct Answer: 6720
Explanation:
There is a 7 - digit telephone number but extreme right and extreme left positions are fixed .
i. e. , 6 x x x x x 5
? Required number of ways = 8 x 7 x 6 x 5 x 4 = 6720
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