Permutations And Combinations Examples

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

C(n,r)= n!/r!(n-1)!

C(12,3) = 12!/3!(12-1)!

C(12,3) = 12!/3!*9!

C(12,3) = 12*11*10*9!/3*2*1*9!

C(12,3) = 220

The answer for the given example problem is 220.

by: Omkar Nayak

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