maths help!

Although 4C2 x 2C1 x 1C1 is totally correct for this example, it is slightly different in terms of the logic we used before. When you solve it like this the intention (still totally valid) is slightly different. Here it is saying. I CHOOSE 2 out of the 4 places for the blue balls. Then from the remaining 2 places, I CHOOSE a place for the red ball. Then from the one remaining place I CHOOSE the place for the green ball. Which is just: 4C2 x 2C1 x 1C1 = 12. This is perfectly fine and if not better logic.

^^ Potentially this is easier to understand as a method rather than my ramblings.

My argument was subtly different. I am saying. I CHOOSE 2 out of the 4 places for the blue balls. Then I find the number of permutations for the red and the green ball in which they can placed in the two remaining places - which is 2P2 = 2 = 2! ways. So my solution is more strictly 4C2 x 2P2 = 12.

We can't do 4P2 as the permutation formula on works for distinguishable objects but here the blue balls are indistinguishable. So 4P2 ends up double counting all the permutations with a blue ball in them. And then after this we don't know if we have placed a blue ball or not. So the next part of our logic may very well break down. This is ultimately why we do the first one as the logic is unclear.

However, if you wished to use permutations for this problem the logic would again be different and the multiplication would look again different.

The main point is which of the permutations and combinations give you the most logical and straightforward approach to the problem. Both can be used to solve most of these problems they just require the correct manipulation of the formulae. And knowing which one is more straightforward to use mainly comes from experience and you'll get into a certain way of looking at the problems.

No - I'm not a guy at all.

I am not anti-Oxbridge but it is the wrong place for some degrees and some people.

Ah thanks a lot - I am really getting this
Now you know how you said"permutation formula on works for distinguishable objects" - this made so much sense!
so I sorta applied that logic into another problem
The other problem I came across was

But here they are doing 3P3 X 5P5 - why not combinations? Like it doesn't really make sense that they used permutations here bc the permuattion formula works for distinguishable objects but there are 3 acute patients and 5 normal here so the pateints fit into 2 categories.

If you get the chance, could you pls lemme know your thoughts on this

Btw again I am so grateful for your help

So here the patients are put into two categories: acute and not acute. However, the patients themselves are distinguishable so we need to use permutations. We know the group of acute patients comes first. So we just need to find the number of ways we can arrange the acute patients and the number of ways we can arrange the non-acute patients. This is 3P3 x 5P5.

Yess!! That's why-thank you very much
I also figured out how to get the answer for this question:

I did 10C2 X 8C2 X 6C2 X 4C1 X 3C3.

But tbh iam not quite understanding the thought process behind this one -why not involve permuattions and why does the method i sorta tried out above work?

Yess!! That's why-thank you very much
I also figured out how to get the answer for this question:

I did 10C2 X 8C2 X 6C2 X 4C1 X 3C3.

But tbh iam not quite understanding the thought process behind this one -why not involve permuattions and why does the method i sorta tried out above work?

The way you have done it numerically implies the thought pattern: I CHOOSE 2 out of the 10 places for the Xs, then I CHOOSE 2 out of the remaining 8 places for the Ys, then I CHOOSE 2 out of the remaining 6 places for the Zs, then I CHOOSE 1 out of the remaining 4 places for the P and then I CHOOSE 3 out of the remaining 3 places for the Qs. Giving 10C2 X 8C2 X 6C2 X 4C1 X 3C3.

A formula you can use is the following. If I have a set of objects with $x_{1}$ objects of type 1, $x_{2}$ objects of type 2, ... and $x_{k}$ of type k. Then the number of possible arrangements of these is $\frac{\left( \sum_{i=1}^{k}x_{i} \right)!}{x_{1}!\times x_{2}!\times \cdots \times x_{k}!}$

So here we can also do $\frac{10!}{2!2!2!1!3!}$ and this gives the same result.

This in fact could have been used with the original problem with the coloured balls.

Another way of thinking about a problem like this is count the number of ways if all the letters were distinct, and then work out how many times you've counted the same combination.

That is, if we add suffixes so your letters are all distinct we get: $X_1X_2Y_1Y_2Z_1Z_2PQ_1Q_2Q_3$, and we now have 10 distinct "letters" with 10! ways of arranging them.

Now imagine we take a particular arrangement, and just concentrate on where the X_i go. There are 2! ways of permuting $X_1, X_2$ and we can't tell them apart. So we need to divide by 2!.

Similarly we divide by 2! for the Y_i, 2! for the Z_i, and 3! for the Q_i. So we end up with $\dfrac{10!}{2!2!2!3!}$.

It's not uncommon with counting arguments to use "tricks" like these, where we change the conditions to get something easier to calculate, but then have to adjust the answer based on the effect of the change.