meager marsh
Finishing up the year in pre-calc we began to cover combinatorics, a few days ago I was practicing permutations and this question popped up, but this wasn't the presented way to solve it- though the answer is correct- as my teacher used either the compliment rule or adding up the possible permutations the long way(using the OR rule). I however, decided to use the fundamental counting theorem and subtract [r • n].
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Finishing up the year in pre-calc we began to cover combinatorics, a few days ag...
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woven elk
Finishing up the year in pre-calc we began to cover combinatorics, a few days ag...
If it's a correct method, it should be provable that it's equivalent to the other method in general.
The way I see it, this is a relatively complex question. You can't allow 0 as the lead digit, and you can't allow all three even digits to be used up before you get to the final digit.
meager marsh
So, if I was to use fundamental counting theorem and subtract [r • n], I should get an equivalent answer when using the complement rule?
If I were to attempt to find, LS = RS, using their generalized formulas?
*if I stated restrictions beforehand
woven elk
One thing, why did you say there are 4 choices for the second digit?
Also, what's your rationale for subtracting r * n?
24?
meager marsh
Oh, I erased it because I thought it was wrong and rewrote the digits from memory, just swap the 5 and the 4
woven elk
...why did you say there are 4 choices for the lead digit?
woven elk
...why did you say there are 3 choices for the second digit?
woven elk
...okay, but then what second number are you not allowing in the lead position?
meager marsh
The end number, one of the even digits was preselected
woven elk
Also, what's your rationale for subtracting `r * n`?
And this?
meager marsh
I was searching for things that would work, that did
I just need to discover if I can use this method by testing it.
woven elk
No, you need to prove it in general, and I expect you would probably fail just because of how, like, haphazard the whole process seems to have been.
meager marsh
I would have to agree, and then attempt it anyway
woven elk
Okay, look, it's a coincidence.
meager marsh
I suspected as such
woven elk
Because let's do it properly, by counting cases. In the case that we reserve 0 for the end digit, we have 5 * 4 * 3 possibilities, right?
In the case that we reserve 2 or 4, we have 4 * 4 * 3 (excluding zero as a lead digit), right?
woven elk
So we have: 5*4*3 + 4*4*3 + 4*4*3 = (5 + 4 + 4)*4*3 = 13*4*3 = 156But also, wholly coincidentally: 5*4*3*3 = 5*4*3 + 5*4*3 + 5*4*3 = 5*4*3 + 4*4*3 + 4*4*3 + 2*4*3 = 5*4*3 + 4*4*3 + 4*4*3 + 6*4
So it's not that it's r*n, it's that it's 2*n*(n - 1), in this specific context.
meager marsh
Alright, though you are over complicating it for a solution it does definitely show where I could get that from(albeit in this specific case)
Thanks
woven elk
...what do you mean, overcomplicating it? I'm making it exactly as complicated as... it is.
It's actually because it's 2*(r - 2)*(r - 3).
meager marsh
Sorry, I mistyped. I mean that showing the expanded factorials was only really necessary to showcase that
woven elk
...what "expanded factorials"?
meager marsh
Hold on, I'll show you how to get the answer using the complement and summative rules
meager marsh
...what "expanded factorials"?
Although it is hardly notable, it does reduce writing and condense it more for on paper(especially for the complement solution)
woven elk
Although it is hardly notable, it does reduce writing and condense it more for o...
Okay, but I'm typing. It would save literally one character per term.
woven elk
So wait, you saw this, and you didn't recognize it as the coincidental equivalent of what you did?
meager marsh
Yep, and I feel really silly about it. Genuinely thanks for helping me out!
meager marsh
So wait, you saw this, and you didn't recognize it as the coincidental equivalen...
Hold on gotta give rep... Thanks!
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