# 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. ### 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

### Steps needed to be followed to solve Combination based Questions:

Here are some strategies to tackle such problems effectively:

1. Understand the Problem: Read the problem carefully and ensure that you understand what is being asked. Identify the number of items to be selected, the size of the set, and any specific conditions or restrictions.

2. Use the Combination Formula: The formula to calculate the number of combinations is $\mathbf{^nC_r = \frac{n!}{(n-r)! r! }}$, where n is the total number of items, r is the number of items to be selected, and “!” denotes the factorial function. Be familiar with this formula and use it when needed.

3. Simplify Factorials: Factorials can become large quickly, especially for larger values of n. Look for ways to simplify factorials by canceling out common terms or breaking them down into smaller parts.

4. Recognize Combinations with Repetition: In some scenarios, combinations might allow repetition of items. For example, choosing a team with multiple members from a group. In such cases, the formula becomes (n+r-1)C(r), where n is the total number of items, and r is the number to be selected.

5. Apply Combinatorial Identities: Familiarize yourself with combinatorial identities like Pascal’s Identity (nCr = (n-1)C(r-1) + (n-1)C(r)) and the Sum of Combination Formula (ΣnCr = 2^n). These identities can help simplify calculations and provide insights.

6. Draw Diagrams or Tables: Sometimes, drawing diagrams or tables can help you visualize the combinations and keep track of your selections, especially in complex scenarios.

7. Divide and Conquer: For problems with multiple steps or conditions, break them down into smaller, manageable parts. Solve each part independently and then combine the results to find the final answer.

8. Look for Patterns: In some cases, patterns may emerge that can help you find combinations without using lengthy calculations. Look for symmetries, repetitions, or regularities in the problem.

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### 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 = (6C1 * 5C3) + (6C2 * 5C2) + (6C3 * 5C1) + (6C4)

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)

5C3 * 3C2 + 5C4 * 3C1 + 5C5

(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= 7C3

Number of ways of selecting 2 vowels from 4= 4C2

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4 = 7C3 × 4C2 = 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 = 8C3

8C3 = $\frac{8!}{(8-3)! 3! }$ = 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 – 1Cr5 + 3 – 1C15 = 7C5

We know that,

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

7C5 = $\frac{7!}{(7-5)! 5! }$= 21

Correct option: A

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