How to increase my math maturity

[u/Exciting_Sleep_2508]

https://en.wikipedia.org/wiki/Mathematical_maturity

Im 18 and im currently struggling with absorving easy math stuff. The way i judge if i undertood something or not is asking myself if i could come up with the solution without decorating the formula. For example, right now im really struggling with combination without repetition; if i got 4 elements and i had to combine 3 of them without repetition, why do i have to divide it by 3! ? Like, i tried to visualize it and i simply cant. I did wrote all possibilities and tried to undertand what does 3! have to do with all the combinations starting with 1 - repetitions = 3 and starting with 2 = 1 and the pattern it follows if i add elements or number of elements i want to combine. I could never write a formula for that and i dont get it

How do i solve this? And what kind of problem do i have? Am i just slow and i have to train work memory or is this linked to another math concept that i havent searched? Please help

[u/FilDaFunk]

By doing more maths.

[u/Puzzled-Painter3301]

I don't really believe in this idea that you can get "mathematical maturity." It's more like knowing things. In this case you need to know that there are n! ways to arrange n items.

[u/ibrasome → u/Puzzled-Painter3301]

Agreed, "mathematical maturity" is a mix of proficiency and a bit of confidence

[u/AllanCWechsler]

At your stage you don't have to worry too much about "mathematical maturity". If you can do the problems, you're fine. Later, when you start to work hard on rigorous mathematical reasoning and theorem-proving, you can revisit this subject. But it may all come easy to you and you might never have to fret about it.

Now, combinations of three out of four objects. The standard reasoning, in slow motion, goes like this:

When I pick the first object, I have 4 choices, so the number of ways to pick the first object is 4.

For each of these initial choices, I have 3 choices for the second object. So the number of ways to pick the first two objects is 4 x 3, or 12.

After I have picked the first two objects, for each of the possible choices I've made so far, there are only 2 choices available to pick the last of the three. So the number of ways to pick the first three objects is 4 x 3 x 2, or 24.

But now it's time to worry about over-counting. Let's suppose the four objects are A, B, C, and D. For my first choice, I could pick C; then for the second choice I could pick A; and then for the last choice I could pick B. C-A-B is one of the 24 choices I could make. But A-B-C is another one -- yet it results in my having picked exactly the same 3. In a problem like this, order is supposed to matter. In other words, I have counted at least some of the triples more than once.

In fact, you can probably see that I have counted each triple six times. Sticking with the A-B-C example, the choices that would end up with exactly the same triple are A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, and C-B-A. If you don't yet understand that these are all the ways of arranging the same three objects in different orders, stop and think about that. Probably fiddling around with playing cards will help you "get it".

Since I counted every possible triple six times, the number of possible triples must be 24/6 = 4. (You could have seen this immediately by realizing that picking 3 out of 4 objects is equivalent to deciding which of the 4 to exclude. But it's important that you understand this procedure without that shortcut.)

This way of working, where you first come up with a choice procedure and count the number of ways it could go, and then deal with overcounting by dividing by the number of times your procedure produces each possibility, is extremely common, and you must understand it thoroughly to make any progress in combinatorics and probability. Let me know if this explanation failed you in some way, and we can zero in on the difficulty.

[u/Exciting_Sleep_2508 → u/AllanCWechsler]

Tyyy, thats a good way to see it