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Combination formula is used to determine the different ways of selecting items from a collection, irrespective of the order of selection. Simply, Combination is the selection of objects from a larger group where order doesn’t matter.
- The formula of combination helps evaluate the number of possible combinations which can be acquired by taking a subset of items from a larger set.
- It represents the different possible subsets that can be formed from the larger set.
The Formula of Combination can be expressed as:
| \({^nC_r} = \frac{{^nP_r}}{r!} = \frac{n!}{r!(n-r)!} \) |
Combinations and Permutations refer to the arrangement of a particular set of data using various method, in various forms, in mathematical reference. In other words, it explains how many times, and in which order a certain set of data can be arranged. The data is basically categorized into sets and subsets. They are also known as the Method of Smart Counting.
Read More: Important Questions For Class 11 Maths Permutations and Combinations
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Key Terms: Combination, Permutation, Binomial Coefficient, Data Set, Combinatorics, Method of Smart Counting, Factorial
What is Combination?
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Combination refers to the act of determining the number of ways of arrangements that can be made from a given set of data, irrespective of the order. Data can be selected in any order. Sometimes it is also called a binomial coefficient.
- Combination, simply, is the combination of “n” things taken “k” at a time without any repetition.
- It can be used for a group of data where the order of data does not matter.
- Combinations and Permutations are often confused with one another.
- However, the differentiating factor between them is that in combination, any order to select data is not needed, while in permutation it does.
- Combinations are mainly used in combinatorics, but they are practised in mathematics as well.
Permutations and Combinations Detailed Video Explanation:
Read Also: Conditional Probability
Solved ExampleExample: A man asks his nephew to choose 4 items from a container. If the container holds 18 items to choose from, how many different answers could the nephew possibly give? (3 Marks) Ans: The given information mentions,
Hence, we have to determine “18 Choose 4” As we know, Combination = C(n, r) = n!/r!(n–r)! Thus, calculating combinations, we get: ⇒ \(\begin{array}{l}\frac{18!}{4!(18-4)!}=\frac{18!}{14!\times4!}\end{array}\) = 3,060 possible answers can be given by the nephew. |
What is Permutation?
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Permutations refer to the number of ways in which a number of objects or subsets can be extracted from a given set of data in a particular order. This means that nb and bn will be considered as two different permutations, while one combination.
- A permutation can be defined as an ordered arrangement of outcomes with an ordered combination.
- The Permutation formula is
| \({^nP_r} = \frac{n!}{(n-r)!} \) |
- Permutations are known to be always greater than the combinations for the given values of n and r.
- In the case of Permutation, both selection, as well as arrangement, are important.
Solved ExampleExample: Name the four-letter words, with or without any legible meaning, that Rana will be able to form from the word “CALM”, without repetition. (3 Marks) Ans: In order for Rana to form four words with four letters without any repetition, the use of permutation is necessary.
Hence, the number of ways by which Rana can arrange the four-letter word can be decided by the factorial of 4: 4 X 3 X 2 X 1 = 24. Hence, Rana will be able to form 24 words from the four-letter word “CALM”. |
Read More: Bayes Theorem Formula
Formula of Combination
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The Formula for Combination is:
| \({^nC_r} = \frac{{^nP_r}}{r!} = \frac{n!}{r!(n-r)!} \) |
The Combination Formula using Permutation can be shown as: C(n, r) = P(n,r)/ r!
Notations nCr Formula
The notions as used can be denoted as:
- r = size of each permutation
- n = size of the set from which elements are permuted
- n and r = non-negative integers
- ! = factorial operator
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Relation between Permutation and Combination
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Permutation and combination formulas come with several similarities.
- Assume that you have n different objects.
- Thus, you have to find the number of unique r-selections (selections which involve r objects) that can be formed from this group of n objects.
- Thus, consider a group of n people.
- In this case, you have to determine the number of unique sub-groups of size r, that can be formed from this group.
- Herein, the number of permutations of the size r is going to be, nPr.
- In the list of nPr permutations, every unique selection is going to be counted r! times, since the given objects in an r-selection can be permuted within themselves in r! ways.
- Hence, the number of unique combinations in this case is: \(\frac{{^n{P_r}}}{{r!}}\)
Hence,
= \(^n{C_r} = \dfrac{^nP_r}{r!} = \dfrac{n!}{(n - r) r!} = \dfrac{n!}{r!(n - r)} \)
Calculating Combinations and Permutations Infograph
How to Apply Combinations Formula?
A combination formula is applied by using factorials and also as per permutations.
- Consider we have “n” objects available, and we want to determine ways in which we can select r things out of these n objects.
- Thus, we first require to find the number of all permutations of the n things considered at r at a time.
- Thus, it would be nPr .
- Now, in nPr permutations, every combination is going to be counted r! times because r things can be permuted amongst themselves in r! number of ways.
Hence, the total number of permutations and combinations of n things, considered r at a time, represented by nCr, will be:
| \(^n{C_r} = \dfrac{^n{P_r}}{r} = \dfrac{n!}{r!(n - r)!} \) |
Read More: Rolle’s Theorem
Things to Remember
- Permutation refers to a number of ways of selecting data or subsets from a given set of lists, the data in the list can be of diverse nature, but the subsets need to be in a particular sequence.
- Combination refers to a number of ways of selecting data or subsets from a given set of lists, where the data can be of diverse nature and no order is required.
- The formula of Combination is \({^nC_r} = \frac{{^nP_r}}{r!} = \frac{n!}{r!(n-r)!} \).
- The formula of Permutation is \({^nP_r} = \frac{n!}{(n-r)!} \).
- Combinations are also known as Binomial Coefficients.
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Previous Years’ Questions
- If nCr−1 = 28, nCr = 56, and nCr+1 = 70, then the value of r is equal to…? [KEAM]
- The number of words that can be formed by using all the letters of the…? [KEAM]
- The number of 4-digit numbers without repetition that can be…? [KCET 2020]
- If nC12 = nC8 then n is equal to…? [KCET 2017]
- A simple graph contains 24 edges. Degree of each vertex is 3…? [KCET 2010]
- How many 5-digit telephone numbers can be constructed using the digits 0 to 9…? [KCET 2014]
- All possible numbers are formed using the digit 1,1,2,2,2,2,3,4,4 taken…? [JEE 2019]
- An eight-digit number divisible by 9 is to be formed using digits from 0 to 9…? [JEE 2014]
- If a, b and c are the greatest values of 19Cp, 20Cq, and 21Cr respectively, then…? [JEE 2020]
- Consider a class of 5 girls and 7 boys. The number of different teams consisting…? [JEE 2019]
Sample Questions
Ques: From 15 participants, 3 are selected winners and to awarded with 10 gift hampers each. In how many ways it could be done? (2 marks)
Ans: Here the n is 15, and k is 3
To find out the number of ways this could be done is – C (n,k)= n!/[(n-k)! k!]
Putting the values in the formula:
= C (15,3) =15! / [15-3)! 3!]
= 455 combinations.
Ques: What are their formulas of Permutations and Combination? (2 marks)
Ans: For selecting r things from n set of things in a particular sequence apply- (n!)/(n-r)!
For selecting r things from n set of things where order doesn’t matter, apply- n! /r! (n-r)!
Ques: Are Permutations and combinations same as factorial? (1 mark)
Ans: No permutations and combination calculations involve calculating factorial, but they are just not similar to factorial.
Ques: How many six digits numbers can be formed from 2,4,6,8 which are divisible by 10 and no other digit is repeated? (2 marks)
Ans: It is necessary for a digit to have 0 at the unit place to make it divisible by 10.
Apart from zero, the digit will contain 5 more random digits. Thus, we need to find out that combination.
So, the number of ways it could be done is 5! = 120.
Ques: Are permutations and combinations the same? (1 mark)
Ans: No, permutation is an arrangement of data in a sequence from a given set, whereas combination is an arrangement of dissimilar data where being in order is not mandatory.
Ques: If given n = 13 and r = 3, find out the number of permutations. (2 marks)
Ans: To know the number of permutations, simply apply the formula and put the values.
(n!)/(n-r)!
= (13!)/ (13-3)!
= (13 x 12 x 11 x 10!)/ 10! = 1716.
Ques: With the same figures, find out the number of combinations. (2 marks)
Ans: For Combination, apply the formula and put the values
= n! / r! (n-r)!
= 13! /3! (13-3)!
= 13 x 12 x 11x10! / 3! (10)!
= 3146
Ques: In how many ways can we choose 5 vowels from the 26 alphabet? (2 marks)
Ans: There would be innumerable ways of choosing vowels from alphabet.
The required number of ways- 26! / 5! (21!)
= 22 x 23x 24x 25x 26/ 5 x 4 x3 x 2 x 1
= 37889280
Ques: How many 4-digit numbers can be formed in between the digits 1 to 9, if the digits aren’t repeated? (2 marks)
Ans: We are here to make sets where order matters, thus we are required to ascertain a number of permutations. The formula used if n= 9 and r= 4 is:
9! / (9-4)! = 9! / 5!
Thus, the final answer for the number of arrangements is: 6 x 7x 8 x 9 = 3024.
Ques: In how many ways can a team of 5 boys and 3 girls be selected from 10 boys and 8 girls? (2 marks)
Ans: 5 boys and 3 girls are to be selected.
To select 5 boys means n=10 and r=5.
To select girls means n=8 and r=3.
Thus, keeping in mind the data, 10! /5! (5!) x 8! /3! (5!) = 210.
Ques: From the English alphabet, with 5 vowels and 21 consonants in total, how many words can be formed from two vowels and three consonants? (2 marks)
Ans: There are 5 total vowels and 2 are to be selected, which means n is 5 and r is 2, then the number of ways the words can be formed are:
5! / 2! x 3! = 10.
Now, there are total 21 consonants and we have to from sets from three consonants. That means n is 21 and r is 3. Given this data, the number of sets to be formed, irrespective of order:
21! / 3! (18!) = 1190.
Ques: How many numbers are there between 99 and 1000, having at least one of their digits 7? (3 marks)
Ans: The numbers between 99 and 1000 are all numbers that come in three-digits.
Total number of three-digit numbers with at least one digit as 7
= (Total numbers of three-digit numbers) – (Total number of three-digit numbers wherein 7 does not occur at all)
= (9 × 10 × 10) – (8 × 9 × 9)
= 900 – 648
= 252
Ques: In a small town, there are around 87 families. Out of the lot, 52 families have around 2 children. As per a recent rural development programme, 20 families were chosen for assistance, out of which almost 18 families ought to have at most 2 children. In how many possible ways the choice can be made? (5 marks)
Ans: As per the given question, it can be said that:
There are:
Total number of families = 87
Number of families with 2 children = 52
The Remaining number of families = 87 – 52 = 35
As given, for the recent rural development programme, 20 families were to be chosen for the sake of assistance.
Here, around 18 families must have at most 2 children.
Thus, the list below is the number of possible choices that can be made:
52C18 × 35C2 (18 families with at most 2 children and 2 selected from other family types)
52C19 × 35C1 (19 families with at most 2 children and 1 selected from other family types)
52C20 (All selected 20 families with at most 2 children)
Therefore, the total number of choices that can be made is = 52C18 × 35C2 + 52C19 × 35C1 + 52C20
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