** Permutation and Combination Aptitude** Questions and Answer – Tips & Tricks

**Factorial:**

Factorial can defined, the product of the number by its successor till it reaches to one.

n! = n (n – 1) (n – 2) ….. 1

**Permutation:**

Permutation, the number of ways particular set can be arranged, when order of the arrangement matters. A combination lock can be called a permutation lock. for example

- We have three letters a, b and c, We have to arrange two letters at a time. So, in this example, the permutations of two letters = ab, ba, bc, cb, ac, and ca.
- If we have to arrange all letters (a,b,c), the permutation would be abc, acb, bac, bca, cab, and cba.

^{n}P_{r} = n(n – 1)(n – 2)(n – 3) … (n – r + 1) = (n! / (n – r)!)

**Combinations:**

The number of ways a particular set can be arranged, where order of the arrangement does not matter which means for a combination of the n number of things there may be different orders. For Example

- Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
- All the combinations formed by a, b, c taking ab, bc, ca.
- The only combination that can be formed of three letters a, b, c taken all at a time is abc.
- Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.
- Note that ab ba are two different permutations but they represent the same combination.

^{n}C_{r} = (n! / ((r!) * (n – r)!)

**Note:**

^{n}C_{n}= 1 and^{n}C_{0}=1^{n}C_{r}=^{n}C_{(n – r)}

Permutation and Combination Aptitude Questions and Answer **With Detailed Explanation** – Quantitative Aptitude

## Permutation and Combination

Question 1 |

5 | |

10 | |

15 | |

20 |

**Answer: Option D**

__Explanation:__

Since, each desired number is divisible by 5, So we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits(2, 3, 6, 7, 9). So, there are 5 ways of filling the ten place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Therefore, Required Number of Numbers = (1 * 5 * 4) = 20

Question 2 |

8 | |

8! | |

7 | |

7! |

**Answer: Option B**

__Explanation:__

Total Number of letters = 8

Using these letters the number of 8 letters words formed is

^{8}P

^{8}= 8!

Question 3 |

30 | |

32 | |

34 | |

36 |

**Answer: Option B**

__Explanation:__

We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear except those that can be selected out of these 6 points are collinear.

Hence, the required number of stright lines

=11C

_{2}- 6C

^{2}- 5

^{2}+ 1 + 1

= 55 - 15 - 10 + 2

**Ans = 32**

Question 4 |

702 | |

624 | |

756 | |

812 |

**Answer: Option C**

__Explanation:__

From a group of 7 men and 6 women, 5 persons are to be selected with at least 3 men.

Hence, we have the following 3 options

We can select 5 men ...(option 1)

Number of ways to do this = 7C

_{5}

We can select 4 en and 1 women.....(option 2)

Number of ways to do this = 7C

_{4}* 6C

_{1}

We can select 3 men and 2 women.....(option 3)

Number of ways to do this = 7C

_{3}* 6C

_{2}

Total number of ways

7C

_{5}+ (7C

_{4}* 6C

_{1}) + (7C

_{3}* 6C

_{2})

7C

_{2}+ (7C

_{3}* 6C

_{1}) + (7C

_{3}* 6C

_{2})

^{n}C

_{r}=

^{n}C

_{(n-r)}

=

= 21 + 210 + 525

**Ans = 756**

Question 5 |

2 | |

8 | |

1 | |

None of these |

**Answer: Option B**

__Explanation:__

When a coin is tossed once, there are two possible outcomes: Head(H) and Tale(T)

Hence, When a coin is tossed 3 times, the number of possible outcomes

= 2 * 2 * 2

**Ans = 8**

Question 6 |

10 | |

20 | |

30 | |

40 |

**Answer: Option B**

__Explanation:__

Number of ways in which the group can be formed

**Ans = 20**

Question 7 |

^{50}P

_{2}

4500 | |

3260 | |

2450 | |

1470 |

**Answer: Option C**

__Explanation:__

Question 8 |

12! | |

10! | |

13! | |

14! |

**Answer: Option C**

__Explanation:__

Number of circular permutations(arrangements) of n distinct things = (n-1)!

Here, n = 10 + 4 = 14

Hence, Number of arrangements possible

=(14 - 1)! = 13!

Question 9 |

9! | |

9! * 3! | |

11! * 3! | |

11! |

**Answer: Option B**

__Explanation:__

Given that 3 particular persons should always be together. Hence, just group these 3 persons together and consider as a single person.

Therefore, We can take total number of persons as 9. These 9 persons can be arranged in 9! ways.

We had grouped 3 persons together. These 3 persons can be arranged among themselves in 3! ways.

Hence, required number of ways

**Ans=9! * 3!**

Question 10 |

210 | |

1050 | |

25200 | |

21400 |

**Answer: Option C**

__Explanation:__

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

=(7C

_{3}* 4C

_{2})

= 210

Number of groups, each having 3 consonants and 2 vowels = 210

Each group contains 5 letters

Number of ways of arranging 5 letters among themselves = 5!

=5 * 4 * 3 * 2 * 1

=120

Therefore, Required Number of ways = (210 * 120) = 25200

List |