Combination
In mathematics, a combination is a way of selecting items from a collection where the order of selection does not matter. Suppose we have a set of three numbers P, Q and R. Then in how many ways we can select two numbers from each set, is defined by combination.
In smaller cases, it is possible to count the number of combinations, but for the cases which have a large number of group of elements or sets, the possibility of a set of combination is also higher. Therefore, a formula has been determined to find the possible selection of the number of items, which we will discuss in this article. We will also discuss the relation between permutation and combination with the help of theorems and their proofs.
Definition of Combination in Math
The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” The combination means “Selection of things”, where the order of things has no importance.
For example, if we want to buy a milkshake and we are allowed to combine any 3 flavours from Apple, Banana, Cherry, and Durian, then the combination of Apple, Banana, and Cherry is the same as the combination Banana, Apple, Cherry. So if we are supposed to make a combination out of these possible flavours, then firstly, let us shorten the name of the fruits by selecting the first letter of their names. We only have 4 possible combinations for the question above ABC, ABD, ACD, and BCD. Also, do notice that these are the only possible combination. This can be easily understood by the combination Formula.
Also, read:
Combination Formula
The Combination of 4 objects taken 3 at a time are the same as the number of subgroups of 3 objects taken from 4 objects. Take another example, given three fruits; say an apple, an orange, and a pear, three combinations of two can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.
More formally, a k-combination of a set is a subset of k distinct elements of S. If the set has n elements, the number of k-combinations is equal to the binomial coefficient.
nCk = [(n)(n-1)(n-2)….(n-k+1)]/[(k-1)(k-2)…….(1)]
which can be written as;
nCk = n!/k!(n-k)!, when n>k
nCk = 0 , when n<k
Where n = distinct object to choose from
C = Combination
K = spaces to fill (Where k can be replaced by r also)
The combination can also be represented as: –nCr, nCr, C(n,r), Crn
Relation between Permutation and Combination
The combination is a type of permutation where the order of the selection is not considered. Hence, the count of permutation is always more than the number of the combination. This is the basic difference between permutation and combination. Now let us find out how these two are related.
Theorem : nPr = nCr.r!
Corresponding to each combination of nCr, we have r! permutations because r objects in every combination can be rearranged in r! ways.
Proof:
nPr = nCr.r!
= [n!/r!(n-r)!].r!
= n!/(n-r)!
Hence the theorem states true.
Theorem: nCr + nCr-1 = n+1Cr
Proof:
\(^{n}C_{r}+ ^{n}C_{r-1}= \frac{n!}{r!(n-r)!}+ \frac{n!}{(r-1)! (n-r+1)!}\)
\(= \frac{n!}{r(r-1)!(n-r)!}+ \frac{n!}{(r-1)! (n-r+1)(n-r)!}\)
\(= \frac{n!}{(r-1)!(n-r)!} \left [ \frac{1}{r} + \frac{1}{(n-r+1)} \right ]\)
\(= \frac{n!}{(r-1)!(n-r)!} \left [ \frac{n + 1}{r (n-r+1)} \right ]\)
\(= \frac{(n+1)!}{(r)!(n+1-r)!}\)
= n+1Cr
Also, learn: Factorial
Example
Example 1: A group of 3 lawn tennis players S, T, U. A team consisting of 2 players is to be formed. In how many ways can we do so?
Solution- In a combination problem, we know that the order of arrangement or selection does not matter.
Thus ST= TS, TU = UT, and SU=US.
Thus we have 3 ways of team selection.
By combination formula we have-
3C2 = 3!/2! (3-2)!
= (3.2.1)/(2.1.1) =3
Example 2: Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} having 3 elements.
Solution: The set given here have 10 elements. We need to form subsets of 3 elements in any order. If we select {1,2,3} as first subset then it is same as {3,2,1}. Hence, we will use the formula of combination here.
Therefore, the number of subsets having 3 elements = 10C3
= 10!/(10-3)!3!
= 10.9.87!/7!.3!
= 10.9.8/3.2
= 120 ways.
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