Combination involves with selection of objects or things out of larger group where order does not matter. Combinations of r objects from a set of n objects can also be done with with repetitions.

**The formulas for Combination are stated as :**

Here,

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Below are the problems based on combinations :

### Solved Examples

**Question 1: **If 10 employees are working in an organization, then how many ways we can arrange a group of 5 individuals each?

** Solution: **

Given total number of employees in the organization = 10 = n

Need to arrange a group of 5 persons in the each group = 5 = r

We know that, Combination = C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{10!}{5!(10 - 5)!}$ = $\frac{10!}{5!5!}$ = 252

**Question 2: **In a competitive exam, each question have 5 different options with 2 correct options. How many ways are there for a applicant to select correct options for each question?

** Solution: **

Given total number of options in the question = 5 = n

For a applicant to select a correct option for each question = 2 = r

We know that, Combination = C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{5!}{2!(5 - 2)!}$ = $\frac{5!}{2!3!}$ = 10

Given total number of employees in the organization = 10 = n

Need to arrange a group of 5 persons in the each group = 5 = r

We know that, Combination = C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{10!}{5!(10 - 5)!}$ = $\frac{10!}{5!5!}$ = 252

Given total number of options in the question = 5 = n

For a applicant to select a correct option for each question = 2 = r

We know that, Combination = C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{5!}{2!(5 - 2)!}$ = $\frac{5!}{2!3!}$ = 10

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