Combinations & Permutations
By DarthVader
Date: 2023-04-11
Topic: 195 see comments
Post views: 1024
Permutations
video: https://www.youtube.com/watch?v=Cg44yLlVCEg
The number of permutations for a set of values is effectively the number of ways it can be re-arranged so the order of the values is different.
For example, if I have a set of five letters: ABCDE, and I want to know how many different combinations of three letters I can make from this, I can use the permutation formula, on the calculator this looks like:
nPr
So to find the answer, I can type in to the calculator:
5(nPr)3 = 60
So there are a possible 60 different ways to arrange the 3 letters from the total of 5 (where each arrangement has the letters in a unique order).
Permutations with repitition not allowed
In this case we would have repeated letters for example: AAABBCDE, but we do not want to count the permutations in which the repeated letters are switched hence giving the exact same arrangement thus ‘overcounting’ the number of unique arrangements of the letters.
In order to calculate the number of unique arrangements or permutations of the letters now, we need to use a different formula:
(P) / (3! × 2!)
where 'P' = the normal permutations calculation i.e. in this example with ‘8’ total letters in the set and we are looking for how many different combinations of ‘3’:
(P) = 8(nPr)3 = 336 (so there are 336 permutations of this set of 8 letters)
Then we must divide this value by the number of repeated letters in the set individually so:
(336) / (3! × 2!)
where 3! accounts for the three letter A's and 2! accounts for the two letter B's.
This gives the answer 28.
So there are a possible 28 unique arrangements of three letters taken from this set of 8 letters.
Permutations with repitition
If in the problem above we did not care about the repeated letters switching places, then we simply need to use the formula:
nr
This tells us that the number of permutations of a 3 letter subset of the 8 letter set [AAABBCDE] is equal to:
83 = 512
Baring in mind that as we are dealing with letters in this case, some of these arrangements calculated in this answer will look exactly the same.
This method is useful though when dealing with numbers for example binary numbers.
The number of possible binary numbers that can be made using a given amount of ‘bits’ is calculated by:
2n
where 2 = the set of 1 and 0 binary and n = the number of bits (binary digits 1/0)
Combinations
Combinations are similar to permutations except that the order of the values does not matter.
On the calculator, the combinations operation looks like this:
nCr
So to find the number of combinations of three letters from a set of five letters, we would type in on the calculator:
5(nCr)3 = 10
So there are a possible 10 different combinations of three letters from the set of 5 (where the order of the letters does not matter).
E.G.
ABC = ACB = BAC = CAB … etc (equals one combination, but 4 unique permutations)
Combinations with repitition
How many ways can ‘r’ objects be selected from a collection of ‘n’ objects with repetition allowed?
To find the number of combinations ‘r’ that can be selected from a set of objects ‘n’ with repeated values allowed the formula is:
(n + r − 1)nPr(n)
Comments | Creator | Date | ID |
---|