**Simple applications (Permutations and Combinations)**

Counting is considered to be one of the foundation stones in mathematics and the most basic things that you would get to learn. The use of permutations and combinations while dealing with larger data would be of great help.

Even of not in its purest and the most accurate forms, the concept of permutations and combinations would be applicable in the simplest task in everyday life for example making a decision for your pizza order.

**Some solved examples for permutations and combinations **

**Question**: A magic show has got about ten people within the audience. In order to perform the next act, the magician requires two people to help from the audience. In how many different ways can he ask two people from the audience to come up to the stage for help?

**Solution**: By the number of ways, what we mean is how many different pairs of people he can invite to the stage. For instance, consider that he has five friends namely, John, Tin, Alice, Robin, and Sarah along with other five people in the audience. The magic trick can be conducted perfectly by inviting Alice and John to the stage and equally well by inviting Robin and Tim to the stage. Thus, we are required to find out the possible number of such pairs that could help in making the magic trick successful.

We are obliged to choose two people out of the total 10 presents as the audience. Thus, as per the formulae we have got r=2 and n = 10. Remember that we are supposed to find the total number of possible combinations. Now,

$^{10}$C$_{2 }$ = $\frac{10!}{2!(10-2)!}$

Expanding the factorial of the number allows solving this further.

$\frac{10.8.2!}{2!.8!}\to \frac{10.9}{1.2}=45$

Therefore, there are 45 ways in which the magician can pick up a pair from the ten people available in the audience.

**Question**: the manager of a sports broadcasting company needs to pick up the best three goals of the month from the available list of ten. How many ways are there to find and decide the top three goals?

**Solution**: Considering the fact that the manager needs the top three goals of the month, it is important to make sure that the chronological order of the goals is maintained. It would help in deciding the first place winner, the first and the second-runner up. Thus, this problem relates to the permutation formula.

In order to pick up three goals from the given list of ten,

Possible permutations i.e. ^{10}P_{3 }= =10 × 9 × 8 = 720.

This means that there are 720 different ways to pick up the top three goals for the month.

The above-mentioned examples would help you well to understand and apply combinations and permutations into realistic problems.