In this article we are going to see:
Simple permutation
Permutation with repetition
Simple disposition
Disposition with repetition
Simple combination
Binomial Coefficients
Considering:
n = elements
k = available positions
Simple Permutation
We had this permutation when there are n possible spaces and n elements.
For example, if I had 7 positions on the podium and 7 horses, every horse will arrive in one of these positions. How can I calculate all the possible combinations?
I can take 7(horses) and calculate the factorial of 7.
That’s mean = 7 * 6 * 5 * 4 * 3 * 2 * 1. The factorial is indicated with !. So 7!.
Why?
Every horse can arrive at every position:
- The first can be occupied by all the seven horses.
- The second can be occupied by six horses because one is already in the first position.
- The third can be occupied by 5 horses because there’s one horse in the first position and one in the second.
- ….
Those are the simple permutation! Pretty easy! For now…
Permutation with repetition
But if I have repeated elements how can I do it?
For example, I have 3 marble, two yellows and one red. The two-position occupied by the two yellow marbles are the same.
How can I calculate in this case?
Doing the factorial of 3, for the 3 marbles, divided by the factorial of every group of marbles, in this case, 2!1! two yellows and one red.
Simple Disposition
When I want to order more elements than the available position. For example, if I want to order 3 elements in seven positions. We do the same things as Simple Permutation but only for the requested positions, in this case: 7 x 6 x 5.
Disposition with Repetition
When the results can be repeated every time, for example, when we throw a die. Every time the possible number is the same, from 1 to 6.
Considering five throws the possible results are 6⁵.
n = 6 (possible elements)
k = 5 (number of lunch)
Simple Combination
In this case, we had to consider a group of elements n.
Every element of the group is different.
So we wanna say all the possible combinations in this group, composed of 8 elements, that contain only 3 elements. Considering that every element is unique, a group can’t exist with the same element, and two groups with the same element but in different positions are the same group. For example, considering:
{a,b,c,d,e,f,g,h} so n = 8
I have to calculate every subset composed of 3 elements.
{a,b,c} and {c,a,b} are the same set.
How can I proceed?
Binomial Coefficients
What’s that?
Count the subset of k beginning with n elements, basically it coun the simple combination, it’s the same formula we have used before.
Don’t you remember what n and k are? Go back to the beginning of the articles!
