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- Question
- The number of different permutations of the word 'BANANA' is?
Options- A. 270
- B. 60
- C. 120
- D. 360
- Correct Answer
- 60
ExplanationThere are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60 - 1. The numbers of straight lines can be formed out of 10 points of which 7 are collinear?
Options- A. 26
- B. 21
- C. 25
- D. None of these Discuss
- 2. From a group of 6 men and 4 women a committee of 4 persons is to be formed?
Options- A. 210
- B. 225
- C. 195
- D. 185 Discuss
- 3. From 4 officers and 8 Jawans in how many ways can 6 be chosen to include at least one officer?
Options- A. 896
- B. 986
- C. 886
- D. 996 Discuss
- 4. In how many different ways can the letters of word JUDGE be arranged so that the vowels always come together?
Options- A. 48
- B. 24
- C. 120
- D. 60 Discuss
- 5. If 7 points out of 12 are in the same straight line, then the number of triangles formed is?
Options- A. 19
- B. 158
- C. 185
- D. 201 Discuss
- 6. Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is?
Options- A. 11
- B. 12
- C. 13
- D. 14 Discuss
- 7. In how many ways can the letter of the word ' civilization' be arranged?
Options- A. 12! / 4!
- B. 12! / 4! -1
- C. 13! / 5! - 1
- D. None of these Discuss
- 8. 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
- 9. 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
- 10. 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
Permutation and Combination problems
Search Results
Correct Answer: 25
Explanation:
If there were no three points collinear. We should have 10C2 lines but since 7 points are collinear we must subtract 7C2 lines and add the one corresponding to the line of collinearity of the seven points.
Thus, the required number of straight lines .
= 10C2 - 7C2 + 1 = 25
Correct Answer: 195
Explanation:
Required no. of ways = [4C1 x 6C3] + [4C2 x 6C2] + [4C3 x 6C1] + [4C4]
= 80 + 90 + 24 + 1 = 195.
Correct Answer: 896
Explanation:
We may choose 1 officer and 5 jawans or 2 officers and 4 jawans ......... or 4 officers and 2 jawans.
So Required answer = [4C1 x 8C5] + [4C2 x 8C4] + [4C3 x 8C3] + [4C4 x 8C2]
= 224 + 420 + 224 + 28 = 896.
Correct Answer: 48
Explanation:
Total number of letters = 5
Number of vowels = 2.
If we consider both vowel as a one letter then,
Required number = 4! 2! = 48.
Correct Answer: 185
Explanation:
The number of ways of selecting 3 points out of 12 points is 12C3. The number of ways of selecting 3 points out of 7 points, on the same straight line is 7C3, Hence, the number of triangle formed will be 12C3 - 7C3 = 210 - 35 = 185.
Correct Answer: 12
Explanation:
Let there be n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n.
nC2 = 66
? n(n - 1) / 2! = 66
? n2 - n - 132 = 0
? (n - 12) (n + 11) = 0
? n = 12
Correct Answer: 12! / 4! -1
Explanation:
There are 12 letters in the world 'civilization' of which four are i's and other are different letters.
? Total number of permutations = 12!/4!
But one word is civilization itself.
? Required number of rearrangements = 12!/4! - 1
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!. ]
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 .]
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!
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