Lets start with definitions.

**Permutation** is the number of subsets you can form from a set considering the **order as a factor** **/ order is important**.

Eg.

set = A,B,C,D

size of set =n= 4

size of subset =k= 2 (pick 2 at a time)

How many subsets of 2 elements can be form?

Order is a factor / important. (this means AB not equal to BA)Answer:

No. of Subsets = AB AC AD BA BC BD CA CB CD DA DB DC = 12 =_{n}P_{k}

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Combination is the number of subsets you can form from a set **without considering the order as a factor** **/ order is not important**.

Eg.

set = A,B,C,D

size of set =n= 4

size of subset =k= 2 (pick 2 at a time)

How many subsets of 2 elements can be form?

Order is not a factor / not important. (this means AB is equal to BA)Answer:

No. of Subsets = AB AC AD~~BA~~BC BD~~CA~~~~CB~~CD~~DA~~~~DB~~~~DC~~= 6 =_{n}C_{k}

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Another form of permutation is the **number of ways a set of numbers can be arranged**. Just like a 4 digits lottery, when you want to buy all the posssible permutations of a set of 4 numbers.

**Eg.**

Number 1234 can be arrange in how many ways?

n = 4

Possible permutations for 1234 = 24**Formula:**

Possible Permutations = n! = 4*3*2*1 = 24

Actually the formula is similar to the one on top. Just that in this case, k = 4.

Therefore:

(n – k)! = (4 – 4)! = 0! = 1**Why 0! = 1?** I will explain that in another post.

What if one of the numbers are repeated.

Like 1231 instead of 1234.

Then the formula n! , will need to be divided by the factorial of the number of repeats.

Possible permutations for 1231 = 4! / 2! = 12

The **2** at the denominator comes from 1, which is repeated **twice**.

What if **2 numbers are repeated**. Like 1221 instead of 1234 or 1231. Then the *denominator will be the product of two factorials(2!*2!) *, since there are 2 numbers which is repeated 2 times each. **The formula: **

possible permutations = 4! / (2!*2!) = 6

Manually listing them:

1221 1122 2211 1212 2112 2121

**A 9 digits example:**

121337999

n = 9, 2 numbers repeated twice , 1 number repeated 3 times. What is the number of permutations?

possible permutations = 9! / (2!*2!*3!) = 15120**If none of the numbers are repeated:**

possible permutations = 9! = 362880