Combinations
Combinations are also called selections. Combinations correspond to the selection of things from a given set of things. Here we do not intend to arrange things. We intend to select them. We denote the number of unique rselections or combinations out of a group of n objects by \(^n{C_r}\).
Combinations are different from arrangements or permutations. Let us learn more about how to calculate combinations, combinations formula, differences between permutation and combinations, with the help of examples, FAQs.
What Are Combinations?
Combinations are selections made by taking some or all of a number of objects, irrespective of their arrangements. The number of combinations of n different things taken r at a time, denoted by \(^n{C_r}\) and it is given by, \(^n{C_r} = \dfrac{n!}{r!(nr)!}\),where 0 ≤ r ≤ n. This forms the general combination formula which is \(^n C_r\) formula.
This formula to find the number of combinations by using r objects from the n objects, is also referred as the ncr formula.
What Is Combinations Formula?
The combinations formula is used to easily find the number of possible different groups of r objects each, which can be formed from the available n different objects. Combinations formula is the factorial of n, divided by the product of the factorial of r, and the factorial of the difference of n and r respectively.
Combinations Formula: \(^nC_r = \dfrac{n!}{r!.(n  r)!}\)
The combinations formula is also referred to as ncr formula. To use the combinations formula we need to know the meaning of factorial, and we have n! = 1 × 2 × 3 × .... (n  1) × n.
Combinations as Selections
Suppose we have a set of 6 letters { A,B,C,D,E,F}. In how many ways can we select a group of 3 letters from this set? Suppose we find the number of arrangements of 3 letters possible from those 6 letters. That number would be ^{6}P\(_3\). Consider the permutations that contain the letters A, B, and C. These are 3! = 6 ways, namely ABC, ACB, BAC, BCA, CAB, and CBA.
Now, what we want is the number of combinations and not the number of arrangements. In other words, the 6 permutations listed above would correspond to a single combination. Differently put, the order of things is not important; only the group/combination matters now in our selection. This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be ^{6}P\(_3\)/3! ways, since each combination is counted 3! times in the list of permutations. Thus, if we denote the number of combinations of 6 things taken 3 at a time by ^{6}C\(_3\), we have:
\(^6{C_3}= \dfrac{^6{P_3}}{3}\). This is also said as 6 choose 3.
A few important results on combinations are as follows:
 The number of ways of selecting n objects out of n objects is:\(^n{C_n} = \frac{{n!}}{{n!\left( {n  n} \right)!}} = \frac{{n!}}{{n!0!}} = 1\)
 The number of ways of selecting 0 objects out of n objects is:\(^n{C_0} = \frac{{n!}}{{0!\left( {n  0} \right)!}} = \frac{{n!}}{{0!n!}} = 1\)
 The number of ways of selecting 1 object out of n objects is: \(^n{C_1} = \frac{{n!}}{{1!\left( {n  1} \right)!}} = \frac{{n \times \left( {n  1} \right)!}}{{\left( {n  1} \right)!}} = n\)
 \(^n{C_r} = ^n{C_{nr}}\)
 \(^n{C_r} +^n{C_{r1}} = ^{n+1}{C_{r}}\)
How To Apply Combinations Formula?
We calculate combinations using the combinations formula, and by using factorials and in terms of permutations. In general, suppose we have n things available to us, and we want to find the number of ways in which we can select r things out of these n things. We first find the number of all the permutations of these n things taken r at a time. That number would be \(^nP_r\) . Now, in this list of \(^nP_r\) permutations, each combination will be counted r! times since r things can be permuted amongst themselves in r! ways. Thus, the total number of permutations and combinations of these n things, taken r at a time, denoted by \(^nC_r\), will be:
\(^n{C_r} = \dfrac{^n{P_r}}{r} = \dfrac{n!}{r!(n  r)!}\)
Relationship Between Permutations and Combinations
Permutation and combination formulas and concepts have a lot of similarities. Suppose that you have n different objects. You have to determine the number of unique rselections (selections that contain r objects) which can be made from this group of n objects. Think of a group of n people – you have to find the number of unique subgroups of size r, which can be created from this group.
The number of permutations of size r will be \(^n{P_r}\). In the list of \(^n{P_r}\) permutations, each unique selection will be counted r! times, because the objects in an rselection can be permuted amongst themselves in \(r!\) ways. Thus, the number of unique combinations can be \(\frac{{^n{P_r}}}{{r!}}\).
\(^n{C_r} = \dfrac{^nP_r}{r!} = \dfrac{n!}{(n  r) r!} = \dfrac{n!}{r!(n  r)}\)
Examples on Combinations
Example 1: Consider the word EDUCATION. This has 9 distinct letters. How many 3letter permutations (words) can be formed using the letters of this word? We now know how to answer questions like this; the answer in this particular case will be \(^9{P_3}\)
Consider the following 3letter permutations formed using the letters A, E, T from EDUCATION:
AET , ATE, EAT, ETA, TAE, TEA
These 6 different arrangements correspond to the same selection of letters, which is {A, E, T}. Thus, in the list of all 3letter permutations, we will find that each unique Combination corresponds to 6 different arrangements. To find the number of unique 3letter selections, we divide the number of 3letter permutations by 6.
Hence, the number of 3letter selections will be \(\dfrac{^9P_3}{6}\) = 60480/ 6 = 10,080
Example 2: Out of a group of 5 people, a pair needs to be formed. The number of possible combinations can be calculated as follows.
\(^5{C_2} = \dfrac{{5!}}{{2!\left( {5  2} \right)!}} = \dfrac{{5!}}{{2!3!}} = \dfrac{120}{2 \times 6} = 10\)
Example 3:The number of 4letter Combinations which can be made from the letters of the word DRIVEN is
\(^6{C_4} = \dfrac{6!}{4!(6  4)!} = \dfrac{6!}{4!2!} = \dfrac{720}{24 \times 2} = 15\)
Important Notes
 Whenever you read the phrase “number of combinations”, think of the phrase “number of selections”. When you are selecting objects, the order of the objects does not matter. For example, XYZ and XZY are different arrangements but have the same selection.
 The number of combinations of n distinct objects, taken r at a time (where r is less than n), is \(^nC_r = \dfrac{^nP_r}{r}\) = \(\dfrac{n!}{r! (nr)!}\)
 This result above is derived from the fact that in the list of all permutations of size r, each unique selection is counted r! times.
 Out of n objects, the number of ways of combinations 0 or n objects is 1; and the number of ways of selecting 1 object or (n  1) object is n.
 Out of n objects, the number of ways of selecting 2 objects is \(^n{C_2} = \frac{{n\left( {n  1} \right)}}{2}\).
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Examples of Combinations

Example 1. In a party of 10 people, each person shakes hands with every other person. How many possible combinations of handshakes can be made?
Solution:
Each unique handshake corresponds to a unique pair of persons. Also, note that the order of the two people in the pair does not matter. For example, if X and Y are two people, then XY and YX won’t be counted separately; only the pair {X, Y} will be counted. Here we consider the combination of any two people who shake their hand.
Thus, we need to find the number of ways in which 2 people can be selected out of 10. Clearly, the answer is:
\(^{10}{C_2} = \dfrac{10!}{2!(10  2)!} = \dfrac{10!}{2!8!} = 45\)
Answer: 45 handshakes in total

Example 2. How many diagonals are there in a polygon with n sides?. (Use the concept of combinations to solve this.)
Solution:
If we select any two vertices of the polygon and join them, we will get either a diagonal or a side of the polygon. Here we can use the concept of combinations. The number of ways of selecting two vertices out of n is \(^n{C_2} = \frac{n(n  1)}{2}\). Out of these selections, n correspond to the sides of the polygon. Thus, the number of diagonals is:
\(\dfrac{n(n  1)}{2}  n = \dfrac{n(n  3)}{2}\).
The number of diagonals are \(\dfrac{n (n  3)}{2}\)

Example 3. A class has 25 students. For a school event, 10 students need to be chosen from this class. 4 of the students of the class decide that either four of them will participate in the event, or none of them will participate. What are the possible combination of 10 students?
Solution:
Given the specified constraint, we divide the set of all possible selections of 10 students into two groups:
 The selections include the 4 students; we already have 4 students – we need to select 6 more students out of the remaining 21 students. This can be done in \(^{21}{C_6}\) ways. Thus, the number of possible selections that include the 4 students is \(^{21}{C_6}\).
 The selections do not include the 4 students; now we need to select 10 students out of the remaining 21. This can be done in \(^{21}{C_{10}}\) ways. Thus, the number of possible selections that do not include the 4 students is \(^{21}{C_{10}}\).
The total number of possible combinations under the specified constraint is \(^{21}{C_6}{ + ^{21}}{C_{10}}\).
FAQs on Combinations
What Are Combinations In Numbers?
Combinations are selections. Selecting r objects out of the given n objects is given by using the factorials. It is denoted by \(^nC_r = \dfrac{n!}{r!.(n  r)!}\). The combinations are the different subgroups that can be formed from the given larger group of objects.
How To Use The Combinations Formula?
Combinations are calculated using the combination formula \(^nC_r = \dfrac{n!}{r! (nr)!}\), while we need to choose r items out of n items. Here we use the factorial formula of n! = 1 × 2 × 3 ×.......(n  1) × n.
Does order matter in combinations?
No, the order does not matter in combination. Just the number of selections or subgroups matters. The number of dresses in the wardrobe can be selected at random in order. Picking 2 clothes out of 8 from the wardrobe requires \(^8C_2\) ways = \(\dfrac{8!}{2! 6!}\) = 28 ways. Here we consider the set of two and do not look into the order of the selection.
What are the possible combinations of 6 numbers?
The possible combinations (selections) out of 6 different numbers are as follows:
 Combination of 1 out of 6 is \(^{6}{C_1}\)
 Combination of 2 out of 6 is \(^{6}{C_2}\)
 Combination of 3 out of 6 is \(^{6}{C_3}\)
 Combination of 4 out of 6 is \(^{6}{C_4}\)
 Combination of 5 out of 6 is \(^{6}{C_5}\)
 Combination of 6 out of 6 is \(^{6}{C_6}\)
What are the possible combinations out of the digits 1234?
The total number of combinations are possible this way: choosing 1 digit out of 4, choosing 2 digits out of 4, choosing 3 digits out of four and choosing 4 digits out of 4 = \(^{4}{C_1}+ ^{4}{C_2} + ^{4}{C_3} + ^{4}{C_4}\) = 4 + 6 + 4 + 4 = 18 ways
How Are Permutations and Combinations Related?
The permutations and combinations are related using the combination formula \(^n{C_r} = \dfrac{^nP_r}{r!} = \dfrac{n!}{(n  r) r!} = \dfrac{n!}{r!(n  r)}\). The combinations are the selection of r things taken from n different things, and permutation is the different arrangement of those r things.
What Are The Differences Between Permutations and Combinations?
Permutations are seen as arrangements of r things out of n things, whereas combinations are seen as selections of r things out of n things. \(^nP_r =\dfrac{n!}{r!}\) and \(^nC_r = \dfrac{n!}{r! (nr)!}\). For the given r things out of n things, the number of permutations are greater than the number of combinations.
What Are the Combination Examples?
The combination examples include the groups formed from dissimilar obects.The formation of a committee, the sport team, set of different stationary objects, team of people are some of the combination examples.
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