VikingWarrior wrote:He was changing the permutation to combination
ie. say ABC was one of the combination of people. We have counted BCA , ACB, BAC, CBA and CAB (total six) in the 10*8*6 but all these refer to the same - a combination of person A, Person B and Person C. To nullify the multiple counting he is dividing it by 6 (which is the number of times each combination repeated in the counting)
Thanks for the explanation but I suppose I am number blind so I still have a doubt...if you want to exclude the additional combinations that are counted then shouldn't you be subtracting something from the 10*8*6...I still don't get how you can divide and why by 6?
Hoping for a clear explanation.
Thanks
Chix began by treating this question as an arrangement question. In other word, order matters.
So, selecting persons A, B, and E is different from selecting E, B and A
Of course, order doesn't matter here (selecting persons A, B, and E is is the same as selecting E, B and A), so within the 10x8x6 arrangements calculated, each unique arrangement is counted more than once. How many times is each unique arrangement counted?
Let's examine the selection A, B, and E.
If we say that order matters, then we will treat the following arrangements as unique:
A, B, E
A, E, B
B, A, E
B, E, A
E, A, B
E, B, A
However, these arrangements are not unique. These 6 arrangements represent the same selection of 3 people.
So, we can conclude that each unique selection has been counted 6 times.
So, to get our answer, we need to divide 10x8x6 by 6
Aside: We can arrange 3 unique things in 3! (6) ways. Similarly, we can arrange 4 unique things in 4! (24) ways. And, in general, we can arrange n unique things in n! ways.
