BASIC Combinations Question (1 Viewer)

[Ilovecarrots]

So I know that 10C2 is the same as 10C8, but why!! Can someone explain to me why this is the case, and also tell me how they will answer this question in an exam.

Thank you

 

[InteGrand]

So I know that 10C2 is the same as 10C8, but why!! Can someone explain to me why this is the case, and also tell me how they will answer this question in an exam.

Thank you

It's because the number of ways to choose 2 objects is exactly the same as the number of ways to choose 8 objects (from a set of 10 objects).

Why? Because for each way of choosing 2 objects, you are also actually choosing 8 objects (and vice versa)! Namely, by choosing 2 objects, you are choosing which 8 objects to leave out.

There's nothing special about 2 and 8 here: using similar reasoning, we have the following general rule:

 

[Ilovecarrots]

It's because the number of ways to choose 2 objects is exactly the same as the number of ways to choose 8 objects (from a set of 10 objects).

Why? Because for each way of choosing 2 objects, you are also actually choosing 8 objects (and vice versa)! Namely, by choosing 2 objects, you are choosing which 8 objects to leave out.

There's nothing special about 2 and 8 here: using similar reasoning, we have the following general rule:

I understand this rule, but I still don't understand the reasoning. Why do we use factorial when we choose 2 and leave 8 out and vice versa? I don't understand why we use factorial, I only apply it.

 

[BenHowe]

The factorial component is because we simply trying to 'select' or 'arrange' a number of elements. When we are selecting elements it is implied we are selecting them together, so the order of each element within a given selection is irrelevant. In actual questions, the reason the order is not important is because the elements are not distinguishable. But that technicality is beyond highschool math.

So you can do an intrinsic first principles approach or you just just use the standard notation. So with the example given say we have 10 balls in a bag and a person will select two at a time. In how many ways can this be done?

First Principles: So consider arranging the 10 balls in a line. We will now partition the line such that we will group two balls and in doing so, group the remaining 8. Now we arbitrarily pick a ball. This ball can be placed in any of the 'positions' in the line, so they are 10 possible positions. The next ball can be placed in any of the remaining 9 positions and so on. This gives 10x9x...x1=10!. However the order in which the balls have been placed is irrelevant due to the reasoning given above. So we must 'undo' the ordering of the balls within the two groups. Thus we divide the total arrangements by 8! and 2!. The order in which this is done is irrelevant since we could have partitioned the line as two balls then 8 balls i.e. _ _ | _ _ _ _ _ _ _ _ or 8 balls then 2 balls i.e. _ _ _ _ _ _ _ _ | _ _



Method 2:

We have 10 balls, we wish to select 2 at a time i.e



Hope this helps