Combinations
Combinations are also called selections. Combinations correspond to the selection of things and not their arrangement. 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 \(^n{C_r}\).
1.  What Are Combinations? 
2.  Combinations as Selections 
3.  How to Calculate Combinations? 
4.  Relationship Between Permutations and Combinations 
4.  Examples on Combinations 
5.  FAQs on Combinations 
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 formula of combinations or the \(^n C_r\) formula.
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 Calculate Combinations?
We calculate combinations by 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
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 selections will 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 selection 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 ways in which this can be done is
\(^5{C_2} = \dfrac{{5!}}{{2!\left( {5  2} \right)!}} = \dfrac{{5!}}{{2!3!}} = \dfrac{120}{2 \times 6} = 10\)
Example 3:The number of 4letter selections 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 selecting 0 or n objects is 1; the number of ways of selecting 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 handshakes are there?
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.
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?
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. 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. In how many ways can the selection of 10 students be done?
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 selections possible under the specified constraint is \(^{21}{C_6}{ + ^{21}}{C_{10}}\).
FAQs on Combinations
What Are Combinations?
Combinations are selections. Selecting r objects out of the given n objects is given by using the factorials. It is denoted by \(^nC_r\).
How Do You Calculate Combinations?
Combinations are calculated using the formula \(^nC_r = \dfrac{n!}{r! (nr)!}\), while we need to choose r items out of n items.
Does order matter in combinations?
No, the order does not matter in combination. Just the number of selections 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.
What are the possible combinations of 6 numbers?
For possible combinations (selections) out of 6 different numbers:
Combination of 1 out of 6 is \(^{1}{C_6}\)
Combination of 2 out of 6 is \(^{2}{C_6}\)
Combination of 3 out of 6 is \(^{3}{C_6}\)
Combination of 4 out of 6 is \(^{4}{C_6}\)
Combination of 5 out of 6 is \(^{5}{C_6}\)
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 = \(^{1}{C_4}+ ^{2}{C_4} + ^{3}{C_4} + ^{4}{C_4}\) = 4 + 6 + 4 + 4 = 18 ways
How Are Permutations and Combinations Related?
Combinations are related to permutations using the formula \(^n{C_r} = \dfrac{^nP_r}{r!} = \dfrac{n!}{(n  r) r!} = \dfrac{n!}{r!(n  r)}\)
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 \(^nC_r = \dfrac{n!}{r! (nr)!}\)