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Permutations and Combinations

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.
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 nPn 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
nPr = 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
nPn = n!
nP1 = 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
nCr = n!/(r! (n-r)!)
Number of shake hands in a group of n persons= nC2= 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.

nC1= n
nC0 = 1 = nCn
nCr = nCn-r
nCr = nPr/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 =7C5 x 3C2
= 7C2 x 3C1
= 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)
= (7C3 x 4C2)
=((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
Therefor,Required number of ways =(210 x 120)
= 25200.

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