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- Question
- In how many ways, 12 balls can be divided between 2 boys, one receiving 5 and the other 7 balls?
Options- A. 1784
- B. 1584
- C. 1854
- D. 1560
- Correct Answer
- 1584
ExplanationHere, order is important, then the number of ways in which 12 different balls can be divided between two boys who receives 5 and 7 balls respectively, is = [12! / 5! 7! ] x 2! = 1584
- 1. There are 10 points in a plan, out of which 5 are collinear. Find the number of straight lines formed by joining them.
Options- A. 36
- B. 45
- C. 30
- D. 35 Discuss
- 2. Find the number of diagonals formed in hexagon .
Options- A. 12
- B. 10
- C. 6
- D. 9 Discuss
- 3. There is a polygon of 12 sides . How many triangles can be drawn using the vertices of polygon?
Options- A. 200
- B. 220
- C. 240
- D. 260 Discuss
- 4. In how many ways, a cricket team of 11 players can be made from 15 players, if a particulars player is always chosen?
Options- A. 1835
- B. 1001
- C. 1635
- D. 1365 Discuss
- 5. In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?
Options- A. 15
- B. 16
- C. 20
- D. 24 Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. O and A occupying end places.
Options- A. 12
- B. 14
- C. 20
- D. 18 Discuss
- 10. E being always in the middle.
Options- A. 18 ways
- B. 24 ways
- C. 48 ways
- D. 20 ways Discuss
Permutation and Combination problems
Search Results
Correct Answer: 36
Explanation:
Required number of straight lines
=nC2 - mC2 + 1
Here, n = 10, m = 5
= 10C2 - 5C2 + 1
= 45 - 10 + 1 = 36
Correct Answer: 9
Explanation:
Hexagon has 6 sides .
? n = 6
? Required number of diagonals = 10C2 - n
= 6C2 - 6
= 6! / [2!(6 - 2)!] - 6 = 6! / (2!4!) - 6
= (6 x 5 x 4!) / (2 x 4!) - 6
= 15 - 6
= 9
Correct Answer: 220
Explanation:
Required number of triangles = 12C3 = (12 x 11 x 10) / 6 = 220
Correct Answer: 1001
Explanation:
Since, particular player is always chosen. It means that 11 - 1 = 10 players are selected out of the remaining 15 - 1 = 14 players.
? Required number of ways = 14C10
= 14 ! / (10! x 4!)
= (14 x 13 x 12 x 11) / (4 x 3 x 2 x 1)
= 7 x 13 x 11
= 91 x 11
= 1001
Correct Answer: 24
Explanation:
Total number of ways = 4C3 x 4C2 = [4! / {3! x 1!}] x [4! /{2! x 2!}]
= (4 x 4 x 3 x 2 x 1) / (2 x 2)
= 4 x 6
= 24
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
Correct Answer: 1023
Explanation:
Required number of ways
= 2n - 1 = 210 - 1 = 1024 - 1 = 1023
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!
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
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
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