Permutations vs. combinations

Thread Starter

boks

Joined Oct 10, 2008
218
What's the difference between these two:

1) The number of permutations of n distinct objects taken r at a time is \[\frac{n!}{(n-r)!}\]

and

2) The number of combinations of n distinct objects taken r at a time is \[\frac{n!}{r!(n-r)!}\]

?
 

hgmjr

Joined Jan 28, 2005
9,029
You have surely already noticed that:

since permutations = \[\frac{n!}{(n+r)!}\]

and combinations = \[\frac{1}{r!} *\frac{n!}{(n+r)!}\]

then combinations = permutations time \[\frac{1}{r!}\]

hgmjr
 

steveb

Joined Jul 3, 2008
2,431
What's the difference between these two:

1) The number of permutations of n distinct objects taken r at a time is \[\frac{n!}{(n-r)!}\]

and

2) The number of combinations of n distinct objects taken r at a time is \[\frac{n!}{r!(n-r)!}\]

?
Can you be more specific as to what you are trying to understand.

The literal answer to your question is that the difference is
\[\frac{n!}{(n-r)!}-\frac{n!}{r!(n-r)!}\]

Surely, that is not what you mean. Are you trying to undrstand what the definitions of permutations and combinations are?
 

Mathematics!

Joined Jul 21, 2008
1,022
permutations - order does matter in counting ABC is different then BCA

combinations - order doesn't matter if you have to choose 2 elements out of 5 elements.

Example {A,B,C,D,E} you don't count {A,B,C} different from {B,A,C}
They are the same

But permutations would consider them both different and count as 2 .
 
To support Mathematics!

combinations = "select"
Ex: In how many ways you can select 3 letters from A,B,C,D,E?
5!/(3!2!)
permutations = "select and arrange"
Ex: In how many ways you can select 3 letters from A,B,C,D,E and then arrange those 3 letters?
5!3!/(3!2!) = 5!/2!
 

jamesprx

Joined Dec 31, 1969
0
Permutation is usually understood to be a sequence containing each element from a finite set once..

Combination is arrangment of elements in any order, to get different possible sequences
 
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