Permutations and combinations probability

Math Permutations and Combinations We have learnt some techniques of counting the objects so that we need not count them one by one. Such techniques have been of interest since thousands of years. We have learnt to solve several counting problems through the fundamental principles of counting and the concepts of permutations and combinations.

Permutation is used when we are interested in different arrangements of the given objects while combination is used when we are interested only in selection of the given objects and it matters little which object is selected first. Note that ab and ba are two different permutations but the same combination. Let us now define permutations and combinations.

Permutation: The word permutation stands for arrangement. An arrangement that can be made with a given number of distinct objects by taking some or all of them is called a permutation.
The notation nPr or P(n, r) stands for permutation of n objects taking r at a time, i.e., arrangements of r objects out of n objects. nPr makes sense only when n > 0, r = 0 and n = r.

Combination: Sometimes we are not interested in arrangement butt only in selection. For example, out of the three persons a, b, c two can be presented in 6 ways; ab, ba, ac, ca, bc, cb. However there will be only 3 combinations: ab, ac, bc. Each of the different groups or selections which can be made by taking some or all of a number of things (irrespective of order) is called a combination.

Calculating permutations and combinations: We will use nPr = n!/(n-r)! for calculating permutations and we use nCr = nPr / r! for calculating combinations . Let us take an example for calculating permutations.
Example: Evaluate 7P3.7P3 = 7!/(7 – 3)! = 7!/4! = (7 × 6 × 5 × 4!)/4! = 210.and if we will Evaluate 5C2  to solve  5C2 = 5! / (2! (5 – 2)!) = 5! / (2! 3!) = (5 × 4 × 3!) / (2 × 3!) = 5 × 2 = 10.

Let us take some examples to understand permutation and combination probability.
Example: One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn.
Find the probability that the card drawn is an ace. Here we will use combination as out of 52 cards, one card can be drawn in 52 C 1 ways.

So, total number of elementary events = 52 C 1 = 52.
There are four aces in a pack of 52 cards, out of which one ace can be drawn in 4 C 1 ways.
So, favourable number of elementary events = 4 C 1 = 4.
So, required probability = 4/52 = 1/13.