I know this is (very) basic, but it's a long time since I did this. What is the mathematical calculation that tells you how many different ways you can compare x objects from a total of y (e.g. how many ways direct 2 way comparisons I could do from a total of 4 objects)...
Vague recollections of nCr or nPr or something like that but nothing forming a definite shape in my head.
Vague recollections of nCr or nPr or something like that but nothing forming a definite shape in my head.
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y choose x, which is: y!/x!(y-x)!
Trying to remember formula does not always help. I would suggest to think about it from scratch.
If you were to compare 2 objects out of a total of 4 you would need to compare the first with the second, then the first with the third, then the first with the fourth. Then you need to compare the second with the third, the second with the fourth. Finally you have to compare the third with the fourth and you're done. This means that you need to make, in the following order, a total of 3, 2 and 1 comparisons. You could make it general by saying the you need (y-1), (y-2) and (y-3) comparisons. Which is simply Sum(y-i) with i moving from 1 to y. Rewrite that formula by taking y out of the Sum symbol. If C is the number of comparisons you are looking for, you get:
C(y)=y-Sum(i)=y-y(y+1)/2=y(y-1)/2
This is only valid for 2 objects. For x objects things get a little bit more complex. You need to use the binomial coefficient, as I wrote above, but you can still get there through reasoning.
I hope this helps.
Trying to remember formula does not always help. I would suggest to think about it from scratch.
If you were to compare 2 objects out of a total of 4 you would need to compare the first with the second, then the first with the third, then the first with the fourth. Then you need to compare the second with the third, the second with the fourth. Finally you have to compare the third with the fourth and you're done. This means that you need to make, in the following order, a total of 3, 2 and 1 comparisons. You could make it general by saying the you need (y-1), (y-2) and (y-3) comparisons. Which is simply Sum(y-i) with i moving from 1 to y. Rewrite that formula by taking y out of the Sum symbol. If C is the number of comparisons you are looking for, you get:
C(y)=y-Sum(i)=y-y(y+1)/2=y(y-1)/2
This is only valid for 2 objects. For x objects things get a little bit more complex. You need to use the binomial coefficient, as I wrote above, but you can still get there through reasoning.
I hope this helps.
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To compare (or choose) r objects from a total of n objects, we need nCr.
nCr = n! / r!(n-r)!
For example, to compare 2 objects from 5, we calculate 5C2 which is
5! / 3!2!
=
120 / 6*2
=
10
nCr = n! / r!(n-r)!
For example, to compare 2 objects from 5, we calculate 5C2 which is
5! / 3!2!
=
120 / 6*2
=
10