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Combinatorics / Pascal's Triangle paths
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Hi there anon,
I need help with this math question.
I've done it both the nCr and pascal's triangle way, and always get 630.
Am I wrong?
Can you help?
Thx.
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>>26117
Want to explain your reasoning a little better?
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630 seems like a reasonable answer. i get that as well.

can't be A or D, since the solution *has* to be divisible by 6. B is too high and C... i have no idea how to get C. btw, what's the deal with points B and C? even if i change the task to go from A through B and C to D i end up with 216...
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>>26135
Sure thing.
>(ignore the formulas; different problem)
I tried twice same result.
Then thought that maybe you have to go through specified points, and did the top left grid, but still didn't get a specified answer.

The nCr calculations are in my calculator, but didn't turn up any useful answers.

Thank you senpai!!
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>>26138
>has to be divisible by 6
Can you help me with the reason for that?
This kind of problem wasn't really explained in the lesson beyond two simple examples.
Thanks for the help.
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>>26140
the most striking thing to me immediately was the bottleneck between C and D. every path has to go through this one single node, allowing to split the problem into two parts. the small square from bottleneck to D has 6 unique paths through it. ergo, however many paths there are to get from A to bottleneck have to be multiplied by 6 to reach the final answer.
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>>26141
I see now!
This is why I was never really that good at things like solving rational equations or trig identities. I can't extrapolate useful stuff like that and turn it back into the question to help solve it.

So I guess I'll choose C, and let you know how it goes in a few minutes.
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>>26141
kek, the next question is numerical response about having to go through B and C.
216 it is.
>Here's hoping.
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>>26138
Well, got them both wrong.
Can you make sense of the explanation?
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>>26153
nah, it's clearly wrong.

the last step before B is a multiplication while literally all other steps are additions (like they're supposed to be). there are no 9 different ways to reach B from A. that's a tiny square, you can fucking count them by hand if the methodological inconsistency isn't enough to alert you.
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>>26117
There are (4 choose 2) ways to get from the bottleneck to D. 4 steps, and choose two of them to be down.

From A:

If you go 2 right, then there are (7 choose 4) ways of reaching the bottleneck.

If you go 1 down and 2 right, or 1 right, 1 down, 1 right, then there are (6 choose 3) ways of reaching the bottleneck.

If you go 2d2r 1d1r1d1r, or 1r2d1r, then there are (5 choose 2) ways of reaching the bottleneck.

The total number of ways is ( (7 choose 4) + 2 * (6 choose 3) + 3 * (5 choose 2) ) * (4 choose 2).

Put it in a calculator and get some integer.
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>>26361
Thanks for figuring that out!
We get 630 again.

I've sent an email to my teacher explaining the error, but he hasn't gotten back to me yet. He did say that others had had had questions about the quiz, so they probably encountered the same problem.
Thank you all for your help.