Permutation and Combination problems
- 1. Letters of the word DIRECTOR are arranged in such a way that all the vowels come together. Find out the total no. of ways for making such arrangement.?
Options- A. 4320
- B. 2720
- C. 2160
- D. 1120 Discuss
Correct Answer: 2160
Explanation:
Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s
Hence no. of arrangement = 6!/2! x 3! = 2160
[3 vowels can be arranged in 3! ways among themselves, here multiplied with 3!. ]
- 2. How many different letter arrangements can be made from the letter of the word RECOVER?
Options- A. 1210
- B. 5040
- C. 1260
- D. 1200 Discuss
Correct Answer: 1260
Explanation:
Possible arrangements are 7!/2! 2! = 1260
[Division by 2 times 2! is because of the repletion of E and R .]
- 3. There are 20 books of which 4 are single volumes and the other are books of 8, 5 and 3 volumes respectively. In how many ways can all these books be arranged on a self so that volumes of the same book are not separated?
Options- A. 7! 8! 5! 3!
- B. 7! 8! 4! 3!
- C. 7! 6! 5! 3!
- D. None of these Discuss
Correct Answer: 7! 8! 5! 3!
Explanation:
Volumes of the same book are not to be separated i, e. all volumes of the same book are to be kept together, Regarding all volumes of the same book as one book, we have only 4 + 1 + 1 + 1 = 7 books.
These seven books can be arranged in 7! ways. The book having 8 volumes can be arranged among themselves in 8! ways, the book having 5 volumes can be arranged among themselves in 5! ways, and the book having 3 volumes can be arranged among themselves in 3! ways.
? Required number = 7! 8! 5! 3!
- 4. A library has two books each having three copies and three other books each having two copies. In how many ways can all these books be arranged in a shelf so that copies of the same book are not separated?
Options- A. 120
- B. 180
- C. 160
- D. 140 Discuss
Correct Answer: 120
Explanation:
Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.
? Required number = 5! x 1! x 1! x 1! x 1! = 120
- 5. 4 boys and 2 girls are to be seated in a row in such a way that two girls are always together. In how many different ways can they be seated?
Options- A. 120
- B. 720
- C. 148
- D. 240 Discuss
Correct Answer: 240
Explanation:
Assume the 2 girl students to be together i,e (one). Now there are 5 students.
Possible ways of arranging them are 5! = 120
Now they (two girl) can arrange themselves in 2! ways.
Hence, total ways = 120 x 2! = 240
- 6. The total number of words, which can be formed out of the letters a, b, c, d, e, f taken 3 together, such that each word contains at least one vowel is?
Options- A. 72
- B. 48
- C. 96
- D. None of these Discuss
Correct Answer: 96
Explanation:
The required number of words is : (2C1 x 4C2 + 2C2 x 2C1) 3! = 96
- 7. A man has 5 friends and his wife has 4 friends. They want to invite either of their friends, one or more to a party. In how many ways can they do so?
Options- A. 9
- B. 18
- C. 31
- D. 46 Discuss
Correct Answer: 46
Explanation:
Number of ways of selecting one or more friends from 5 friends
= 5C1 + 5C2 + 5C3 + 5C4 + 5C5
= 5 + 10 + 10 + 5 + 1
= 31 ways
Number of ways of selecting one or more friends from 4 friends = 4C1 + 4C2 + 4C3 + 4C4
= 4 + 6 + 4 + 1
= 15 ways
? Total number of ways = 31 + 15 = 46 ways
- 8. If there are 10 positive real numbers n 1 < n 2, n 3 .... < n 10 How many triplets of these numbers (n 1, n 2, n 3) (n 2, n 3, n 4), ... can be generated such that in each triplet the first number is always less than the second number and the second number is always less the third number?
Options- A. 45
- B. 90
- C. 120
- D. 180 Discuss
Correct Answer: 120
Explanation:
Three numbers can be selected and arranged out of 10 numbers in 10P3 ways 10!/7! = 10 x 9 x 8
Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways.
Hence, required number of arrangement = (10 x 9 x 8)/(3 x 2)
= 120 ways
- 9. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?
Options- A. 990
- B. 2730
- C. 12870
- D. 1560000 Discuss
Correct Answer: 12870
Explanation:
Total number of password using all alphabets -Total
number of password using no symmetric alphabets
= (26 x 25 x 24 ) - (15 x 14 x 13 )
= 12870
- 10. A question paper had 10 questions. Each question could only be answered as True (T) or false (F). Each candidate answered all the questions. Yet, no two candidate wrote the answers in an identical sequence. How many different sequences of answers are possible?
Options- A. 20
- B. 40
- C. 512
- D. 1024 Discuss
Correct Answer: 1024
Explanation:
Each question can be answered in 2 ways.
? 10 question can be answered = 210 = 1024 ways
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