What is Permutation and Combination?

Permutation(P) is a term associated with "arrangement" of alike or different objects according to any rule (if any) set by us at a particular time. For example, "find the permuation of 3 balls red,black and blue"

Combination(C) means the way of selecting one or more terms from a specific set of objects. Generally, combination is associated with "selection". In combination, arrangement of objects in any particular order does not matter as we are just selecting. For example, in how many ways can we select 2 balls from a bag of 5 balls?

Basically, permutation and combination deals with the principle of "advanced" counting.

Permutation or Arrangement of r objects from a set of n objects is shown as _(n,r)P or _rPⁿ or n!/(n-r)!

where k! represents factorial of k, that is, 1*2*3.......*k

Note: 0!=1

Now, permutation or arrangement of n objects=n!/(n-n)!=n!/0!=n!=n*(n-1)*(n-2)* ....1

Combination or Selection of r objects from a set of n objects is shown as _(n,r)C or _rCⁿ or n!/{(n-r)!*r!}

Selection of n objects=n!/{(n-n)!*n!} = n!/(n!*0!)=n!/n!=1

Case 1 of Permutation and Combination

When we have to include particular "x" things among r things from total n things, the number of combination becomes

r-xC^n-x

Relation between Permutation and Combination

_rPⁿ = (_rCⁿ)*(r!)

Case 2 of Permutation and Combination

When we have to arrange N things such that r things among them are equal, number of arrangements are:

N!/r!

Similarly, if we have to arrange N things out of which x are alike, y are alike and goes on, the number of arrangements are shown as:

N!/(x!*y!* ...)

Circular Permutations

Circular permutations are a special kind of permutation.

Number of circular permutations for n distinct objects is given as (n-1)!

If we consider clockwise and counter-clockwise permutation to be same, then the number of permutations is given by (n-1)!/2

Note.

Permutation of n different objects taken r at a time, when repetition is considered is shown as n^r

Case 3

Number of ways atleast 1 object can be selected from p alike objects, q alike objects of second type, and r alike objects of third type is given by

{(p+1)*(q+1)*(r+1)}-1

Case 4

Number of ways atleast 1 object can be selected from p alike objects, q alike objects of second type, r alike objects of third type, and rest different is given by

{(p+1)*(q+1)*(r+1)*(2^n-(p+q+r))}-1

Case 5

Number of distinct solutions of x₁+x₂+ ...x_r=n is given by _r-1C^n-r+1

Now, if we want only positive solutions, all we have to do is add one to every term to make sure it is positive and then find out the cases, that is

(x₁+1)+(x₂+1)+(x₃+1)......+(x_r+1)=n

That is, x₁+x₂+x₃...+x_r=n-r

Thus, number of cases= _r-1C^n-r-r+1

Some useful tips

• If _aCⁿ = _bCⁿ, then either a=b, or a+b=n
• _rCⁿ + _r-1Cⁿ = _rCⁿ⁺¹
• ₀Cⁿ+₁Cⁿ+........_nCⁿ=2ⁿ