Permutations and Combinations
2. Fundamental Principle of Counting
2.3. Combinations
A combination is a grouping or selection of all or part of a number of things without reference to arrangement of the things selected. Selection of three members from a family of ten, number of cricket matches in a group having 10 teams, number of triangles formed by joining any three points out of 20 points in plane etc. all are example of combination. The symbol \({}^n{C_r}\) represents total number of selections taking r different objects out of n different objects.
\({}^n{C_r}\) =\(\cfrac{{n!}}{{(n - r)!(r!)}}\)
=\(\cfrac{{{}^n{P_r}}}{{r!}}\)=\(\cfrac{{n(n - 1)(n - 2)...3.2.1}}{{r!}}\)
Suppose there are 5 students and a company wants to select any 3 of them. So the total number of ways = selection of 3 out of 5 = rejection of 2 out of 5.
Number of ways = 5C3 = 5C2 or in general,
\({}^n{C_r}\) = \({}^n{C_{n - r}}\)
\(^{n + 1}{C_r}{ = ^n}{C_r}{ + ^n}{C_{r - 1}}\)
Suppose there are n + 1 students and we have to select r students out of these students. Now there are two cases:
Case 1: A particular student X is always selected, total number of ways of selecting remaining r – 1 students from the remaining n students =\(^n{C_{r - 1}}\)
Case 2: A particular student X is never selected, total number of ways of selecting r students from the remaining n students =\(^n{C_r}\). Hence total number of ways =\(^n{C_{r - 1}}\)+\(^n{C_r}\).
Thus \(^{n + 1}{C_r} = {\;^n}{C_r}\,\, + \,{\,^n}{C_{r - 1}}\)
Example 01: There are 20 non collinear points in a plane. How many distinct triangles can be formed by joining any three points?
We know that by talking any three points, we can make a triangle, if the points are non collinear. Thus the number of triangles will be \(^{20}{{\rm{C}}_3}\) = 1140
Example 02: In how many ways we can make two groups of 10 and 3 students out of a total 13 students.
We can select either 10 or 3 out of 13. Hence answer is