Hey Developer’s, I’m back with a new topic which is Permutations in the series of statistics foundations.

### Quick Refresher

Counting Sample Points: It means how many sample points are there in the sample space.

So, let’s get started …

### Permutations

Given a set of elements, all distinct arrangements of these elements are called the permutation of a set.

The number of ways you can arrange N elements out of N possibilities is…

P(n,n) = N X (N-1) X (N-2) X …. X 2 X 1 = N! permutations

N! = “N factorial” = a product of all numbers from N down to 1

• Assumes N is a non-negative integer
• By definition,0! = 1

General Formula for Permutations : n!/(n-k)!

Suppose there are n= 5 balls (RGBWO) and you draw k = 3 without replacement. How many possible arrangements are there ?

Solution : 5 possibilities on first draw, 4 on the second, 3 on the third, so 5 X 4 X 3 = 60.

But how to express it in terms of n and k ? : 5!/(5-3)! = 5!/2! = 60

Thus, the number of permutations of k elements out of n is P(n,k) = n!/(n-k)!

Example : Number of unique 5-letter words that can be arranged from letters {a,b,c,d,e} : 5**5 = 3125 (With replacement)

If each letter appears exactly once : 5! = 120 (Without replacement )

3-letter words, each letters appears once : P(5,3) = 5!/(5-3)! = 60

#### Next Post will be on Combinatorics

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