- Home
- Allegations and Mixtures
- AP GP HP
- Arithmetic Progressions
- Averages
- Boats and Streams
- Geometric Progressions
- Harmonic Progressions
- Clocks
- Calendar
- Clocks and Calendars
- Compound Interest
- Simple Interest
- Simple Interest and Compound Interest
- Linear Equations
- Quadratic Equations
- Co-ordinate geometry
- Perimeter Area Volume
- Divisibility
- HCF and LCM
- HCF
- LCM
- Number System
- Percentages
- Permutations Combinations
- Combinations
- Piipes and Cisterns
- Probability
- Work and Time
- Succesive Discounts
- Heights and Distance
- Decimals and Fractions
- Logarithm
- Venn Diagrams
- Geometry
- Set Theory
- Problem on Ages
- Inverse
- Surds and Indices
- Profit and Loss
- Speed, Time and Distance
- Algebra
- Ratio & Proportion
- Number, Decimals and Fractions

- Prepare
All Platforms Programming Aptitude Syllabus Interview Preparation Interview Exp. Off Campus - Prime Video
- Prime Mock

- Interview Experience
- Prime VideoNew
- Prime Mock
- Interview Prep
- Nano Degree
- Prime Video
- Prime Mock

# How To Solve Combination Questions Quickly

## How to solve Combination Problems Quickly

How to solve Combination Questions Quickly is a frequently asked question on google, and we will be discussing the same thing over this page.

**Combination Questions are mostly on the basis of finding the number of ways of selecting the items .**

**Questions on Combination can be solve by formula,**\mathbf{^nC_r = \frac{n!}{(n-r)! r! }}

### How to solve Combination Questions Quickly & Definitons

- Combination is an arrangement of objects where order does not matter.
- There are two easy methods for solving combination questions
- Repetition are allowed
- Repetition are not allowed

### Type 1: How to Solve Combination Question (with or without repetition)

**Question 1.There was a flock of sheep in which 6 were male sheep 5 were female sheep. Now we need to select 4 sheep to take out wool from them. In how many different ways can they be selected such that at least one male sheep should be there? **

**Options:**

A. 1450

B. 302

C. 295

D. 154

**Solution: **The selection can be made in following manner

(1 male sheep and 3 female sheep) or (2 male sheep and 2 female sheep) or (3 male sheep and 1 female sheep) or (4 male sheep)

Required number of ways = (_{6}C_{1} * _{5}C_{3}) + (_{6}C_{2} * _{5}C_{2}) + (_{6}C_{3} * _{5}C_{1}) + (_{6}C_{4})

Required number of ways = (6 * 5) + (15 * 10) + (20 * 5) + 15

Required number of ways = 30 + 150 + 100 + 15

Required number of ways = 295

**Correct option: C**

**Question 2.Among a set of 5 white balls and 3 blue balls, how many selections of 5 balls can be made such that at least 3 of them are white balls.**

**Options:**

A. 45

B. 46

C. 44

D. 40

**Solution: **The selection can be made in following manner

(3 white ball and 2 blue ball) or (4 white ball and 1 blue ball) or (5 white ball)

_{5}C_{3} * _{3}C_{2} + _{5}C_{4} * _{3}C_{1} + _{5}C_{5}

(10 * 3) + (5 * 3) + 1

30 + 15 + 1

46

**Correct option: B**

### How To Solve Quickly Combination Questions

**Question 3. There are 7 consonants and 4 vowels. Find out how many words of 3 consonants and 2 vowels can be formed?**

**Options: **

A. 120

B. 102

C. 20

D. 210

**Solution: **Number of ways of selecting 3 consonants from 7= _{7}C_{3}

Number of ways of selecting 2 vowels from 4= _{4}C_{2}

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4 = _{7}C_{3} × _{4}C_{2 }= 35 * 6 = 210

**Correct option: D**

**Question 4. In how many ways can a team of 5 cricketers can be formed out of a total of 10 cricketers such that two particular cricketers should be included in each team?**

**Options: **

A. 66

B. 65

C. 56

D. 22

**Solution:** Two particular cricketers should be included in each team.

Therefore, select remaining 5-2=3 cricketers from 10-2=8 cricketers = _{8}C_{3}

_{8}C_{3} = \frac{8!}{(8-3)! 3! } = 56

**Correct option: C**

**Question 5. There are 3 types of t-shirts available in a clothing store. In how many ways can 5 t-shirts be selected?**

**Options:**

A. 21

B. 42

C. 5

D. 12

**Solution**: ^{r + n – 1}C_{r} = ^{5 + 3 – 1}C_{15} = ^{7}C_{5}

We know that,

^nC_r = \frac{n!}{(n-r)! r! }

^{7}C_{5} = \frac{7!}{(7-5)! 5! } = 21

**Correct option: A**

**Read Also** –

Login/Signup to comment