subject: Permutations And Combinations Examples [print this page] Introduction of permutations and combinations:
An r-permutation of n items is a sequence structured (arrangement) selection of r items from a set of n items.The number of r-permutations of n items is denoted by P(n,r)
A combination is a group of items (in non-sequence order - selection) from a known set. An r-combination of an n-set S is an r-subset of S. We indicate by (n r) the number of r-combinations of an n-set, read "n choose r." and is denoted by n C r
Solving Permutations and Combinations
Permutation:
An n-C of n objects is also said to be a permutation of n objects. The number of permutations of n
objects is denoted by n!, read n factorial.
The number of r-permutations of n-set equals:
P(n,r) = n(n-1)....(n-r + 1) = n!/(n-r)!
Once looking at permutations of n distinct items we locate it helpful to begin the number of permutations of n separate items taken r at a point as follows:
Assume that n people show up at a theatre but inside the main doors there is only room for a queue of length r people between the doors and the ticket window. The number of such queues of span r is able to follow by filling in the spots one at a time. Let us consider that There are n choices for the first spot, (n-1) for the second... and (n-r+1) choices for the rth spot (note that after the rth spot is filled, (n-r) people remain outside so that at the previous step there were (n-r+1) people to choose from). The permutation is,
n P r = n(n-1)(n-2)...(n-r+1) which can be written as
n P r = n!/(n-r)!
Combination:
r-combination of an n-set equal:
n C r = n!/r!(n-1)! = P(n,r)/r!
In combination the entire number of ways of queues is n C r times r! But this must to be n P r as
we've already seen, i.e.
n C r * r! = n P r, or n C r = n P r /r!, equivalently
n C r = n!/r!(n-r)!
Examples of Permutations and Combinations
Permutation example:
Question: How many different diagrams can be made by 5 lines from 8 lines of different colours?
Solution:
Number of ways taking 5 out of 8 lines.
Using the permutation formula we get
8P5 = 8! / (8-5)!
= 8! /2!
= 8 x 7 x 6 x 5 x 4 = 6720
The answer for the given example problem is 6720
Combination example:
Question: In how many ways can you select a committee of 3 people from a group of 12 members?
Solution: here we need to use combination because the 3 person committee are not ordered
