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CA Foundation Exam June 2023 » CA Foundation Study Material » Business Mathematics » Combinations with standard Results
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Combinations with standard Results

The concept of permutation and combination is a crucial topic in business mathematics which deals with several subtopics like collection and choices selection. In this article we will discuss the concept of permutation and combination with several examples.

Table of Content
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Introduction

Let us assume a scenario to understand the concept of permutation and combination. What if someone asks you to draw 4 cards from a well shuffled pack of 52 cards. Now you can select and draw 4 cards and suppose you note down the cards that you drew.

Now if that person asks you to draw 4 cards again from a well shuffled pack of remaining cards and asks you what is the case that the cards drawn in both the cases are the same.

The study of these types of combinations and the chance of this selection is known as the study of combinations with standard results.

What is a Combination?

The combination helps us to guide the number of ways by which we can choose several objects. In other words, each of the choices which can be made by taking some or all of the number of the things is a combination. The collection or the choices of things are, regardless of the order of choice or selection. 

Let’s take an example, imagine someone asks you to choose or draw four cards from a pack of well-shuffled 52 cards and, by picking up the card, you remember the card and note it down. Then in another round again you shuffled the cards and drew the 4 in a total number of cards.

The cards drawn in both cases may be or may not be the same. The real part in drawing the card lies in drawing the number of ways of drawing the cards, besides it does not interest you if in which round of draw which card was drawn.

The number of ways in which choosing or drawing the card or selecting anything of your choice or interest is the idea of combination.

On the other hand, there is Permutation, Permutation is a mathematical technique that calculates or determines the numbers of manageable arrangements in a set when the order of the contents of the arrangement.

Permutation can be represented by the formula mentioned below:

P(n, k) = n!/(n – k)!

Where, 

  • n is the total number of components in the set.
  • k is the number of selected components arranged in a specified order 
  • ! is factorial

Formula of Permutation and Combinations

Now that you have understood the concept of permutation and combinations drawing standard results. You may be wondering how to draw conclusions from a set of combinations.

Well to understand that here is a list of formulas for permutation and combinations with several examples to help you understand the concept of combinations with standard results in a better way.

Let us suppose if there is a data which is represented by an and there is a choice of X things then the permutation can be defined and calculated by the formula.

nPr =(n!)/(n−x)!

 

Now let’s suppose if there is a group of n data available and you have to select X things without considering the order and replacement then the formula to calculate the combinations available to bring out standard results is

nCr = (nr) =(nPr)/r!=(n!)/r!(n−r)!

Examples of Combination and Permutation Problems

Till now we have discussed the concept of permutation and combination and you have become familiar with the formulas used to calculate the permutation and combination to draw out specific results from a given set of problems. Let’s take a look at some of the solved examples of permutation and combination word problems.

calculate the number of combinations and permutations available if the data is equal to 14 and you have to pick three choices.

As given in the question that data i.e n=14

And the choices, r=3

 

By using the permutation formula we have

nPr =(n!)/(n−r)!=(14!)/(14−3)!=(14!)/(11!)=14×13×12×11!/11!= 2184

 

And the number of combinations can be calculated by using the combination formula as

nCr = (nr) =(nPr)/r!=(n!)/r!(n−r)!=(14!)/(14−3)!=14!/3!(11!)=14×13×12×11!=1092

Conclusion

In mathematics, permutations and combinations are a great way to solve the problems consisting of selections and choices. The concept of combinations is a good and scoring topic if you are preparing for competitive exams that include basic mathematics. In this article we have discussed the concept of combinations drawing out specific results with the formulas and some solved examples. 

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