**Factorial:**

**Factorial notation:** n!=n(n-1)(n-2)(n-3)…….3.2.1

eg: 0!=1 , 1!=1 , 2!=21

Factorial of Fraction & negative integer are not defined.

n!=n.(n-1)!

(2n)!=2^{n}.n!

If the order does matter it is a Permutation.

If the order doesn’t matter, then it’s a combination

**Permutation : **

The different arrangements which can be made out of a given objects by taking some or all at a time.The no. of permutation of n distinct objects into n distinct places is denoted as ^{n}P_{n} read as permutation of n distinct objects taken n at a time.

The number of Permutation of n distinct objects taken from a group of r distinct objects ,in n different places

^{n}P_{r} = n!/(n-r)!

The number of Permutation of n different thing in a circle taken together = (n-1)! But if distinction can’t be made b/w clockwise and anticlockwise arrangement , the no. of arrangement will be halved

(n-1)!/2

^{n}P_{n} = n!

^{n}P_{1} = n

**Combination : **

The different groups which can be made out of given number of objects by taking some or all at a time irrespective of their order.Combination is also known as selections

Number of combinations of r distinct things taken out of n distinct things is given by

^{n}C_{r} = n!/(r! (n-r)!)

Number of shake hands in a group of n persons= ^{n}C_{2}= n(n-1)/2

Order is not important in case of combinations Thus if 4 distinct Alphabets Such as A,B,C&D then

no. of permutation of any 3 alphabet out of is given by

ABC, ABD, ACB, ADB, ACD, ADC, etc whereas no. of combinations is ABC,ABD,ACD.

^{n}C_{1}= n

^{n}C_{0} = 1 = ^{n}C_{n}

^{n}C_{r} = ^{n}C_{n}-r

^{n}C_{r} = ^{n}P_{r}/r!

**Factorial Table :**

**For Example-**

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

**Explanation :**

We need to select 5 men from 7 men and 2 women from 3 women

Number of ways to do this =^{7}C_{5} x ^{3}C_{2}

= ^{7}C_{2} x ^{3}C_{1}

=((7*6)/(2*1))*3

= 21 x 3

= 63

**For Example- **

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

**Explanation :**

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (^{7}C_{3} x ^{4}C_{2})

=((7 x 6 x 5/3 x 2 x 1)*(4 x 3)/(2 x 1))

= 210

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! =5*4*3*2*1

=120

Therefor,Required number of ways =(210 x 120)

= 25200.