Permutation and Combination


In permutations, we saw the number of arrangements. But what if we only want to know the number of ways in which we can choose a given number of objects from a bigger set? In that case, we use combinations. Combinations represent the selection of some or all of different objects in which order of selection doesn’t matter. So, how can we find the number of combinations? Also, what is the combination formula? Let us see in the below space!

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The number of combinations is the number of ways in which we can select a group of objects from a set. For example, if you have ‘n’ objects, in how many ways can you select or choose these ‘n’ objects? Moreover, if the order is taken into consideration then it is the same as the number of permutations. But since the order doesn’t matter, there is only one way to do it! Which means that if you have to select ‘n’ objects taking ‘n’ at a time, there is only one way to do it.

Combination Formula

How about something smaller than ‘n’? Let us see this with the help of an example. Consider that there are 4 objects and you have to select 2 objects from them. Then how many selections can you do? You can pick the first two, the second two, the middle two, the first and the last and so on. If you count, you will find that there are exactly 6 ways to do it.

Thus, combinations are just permutations where the order is not taken into account. So the number of permutations will always be greater than the number of combinations. Using the definition of permutations, we can get the combination formula. Let us see how!

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Combination Formula

Let us say that we have 10 items out of which we will have to select 2 items. How many arrangements can we make? The number of arrangements will be given by = 10P2 = 90. So there are 90 arrangements that we can make from 10 objects if we take 2 at a time. What if the order of the arrangement was not taken into account? For example, we mark one object A and the other B. Then if AB and BA are considered as one arrangement, we say that order doesn’t matter. What will be the number of arrangements in such a case? In that case, it will be the number of ways we can select two items out of a group of 10 items.

To get that, we need to cancel the number of arrangements that are generated because of order. For example, if we take 2 objects then they can be arranged in 2 factorial(2!) ways and so on. So we need to cancel these 2 factorial ways. Thus the number of ways in which we can “select” 2 items from a group of 10 items = 10P2 /2!. This is the combination formula.

In general, we say that if we have a group of ‘n’ objects out of which we make a selection taking ‘r’ objects at a time, then the number of such selections or arrangements is given by nPr/r!

This is known as the combination formula. We represent combination formula as nCr = n!/r!(n-r)!

Other names for it are ‘n choose r’ or ‘binomial coefficient’.

Let us see real-world applications of combination formula.

Solved Examples For You

Example 1: Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} having 4 elements.

A) 340 B) 430 C) 330 D) 430

Answer: C. The set has 11 elements. Any subset that we form has to have 4 elements from the set. Here the order of choosing the elements doesn’t matter. The set { 1, 2, 3, 4} is the same as {4, 3, 2, 1}. Therefore, this is a problem in combinations.
We can do this by using the combination formula as:

11 C 4 = 11!/4!(11-4)! = 11!/7! = ( = 330 ways.

Example 2: The Indian Cricket team consists of 16 players. It includes 2 wicketkeepers and 5 bowlers. In how many ways can you select a cricket team of eleven players if you have to select 1 wicketkeeper and at least 4 bowlers?

A) 1024 B) 1028 C) 1092 D) 1084

Answer: C. If we have to select a team of 11 players from a roster of 16 players then the total number of ways would be 16C11. But here we have to select 11 players including 1 wicketkeeper and 4 bowlers or 1 wicketkeeper and 5 bowlers.

Note that there are a total of 2 wicketkeepers and 5 bowlers to choose from. So the number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players =

2C1×5C4×9C6 = 840.

Furthermore, the number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players.

2C1×5C5×9C5 = 252

Therefore, the total number of ways of selecting the team = 840 + 252 = 1092.

Practice Questions

Q 1: From a group of 6 men and 4 women we have to choose a committee of 5 people. How many committees are possible if there are no restrictions?

A) 262 B) 252 C) 785 D) 9800

Ans: B) 252

Q 2: From a group of 6 men and 4 women we have to form a committee of 5 people. How many committees are possible if there are to be 3 men and 2 women?

A) 130 B) 120 C) 140 D) 180

Ans: B) 120

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2 responses to “Factorial Notation”

  1. Vijayalaxmi says:

    Sir cards probability questions

  2. mk Awas says:

    How to Learn Factorial notation

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