Recent content by cronuscronus

That's a better way of explaining what I was trying to express :). Thanks!

Right, the answer there is 25. So the only thing I can really think of is that we have 5 bins and in each bin there's a maximum of 2 coins that can't be assigned to that bin, so we have to subtract 5^2 from 1001. Is that on the right track?

Using a small script I wrote I see the answer for c should be 976, but I need to think about why that is. :)

I apologize. I'm new to the forums. I added the question to the body as well. Thanks

Hi all. I was hoping somebody could help me with my reasoning here? This is a twist on a typical balls and bins problem. How many ways are there to pick a collection of 10 coins from piles of pennies, nickels, dimes, quarters, and half-dollars? Base on the following condition: a) Assuming that...

I am going to explain to a friend tonight. If you're ever in San Diego I'll buy you a beer :)

It looks like we need to calculate (A to B * B to D) - (B to C * C to D * A to B) so (35*3003)-(21*56*35) = 63,945 routes for part d.

I think part d has an error. We need to calculate (A to B * B to D) - (B to C * C to D), which equals (35*3003) - (21*56) 105105 - 1176 = 103,929

Thanks again for the help. For c, we want to know how many routes pass through both B and C. I think this is more straight forward. Can simply multiply our way through this. A to B is 35. For B to C, we have 5 down, and 2 right required to get to C. So, we can calculate 7choose2, which equals...

Thanks for sticking with me. So, A to D is 497420 options. A to B is 35 options. B to D is 10 down, 5 right. 15choose5 = 3003. Now we want to know how many routes through B, so we can "avoid" that. The routes through B is (35 * 3003). Which equals 105,105. So, 497420-105105 = 392,315...

So I drew this out, and you're correct. You would want to multiply, not add. So if we could get from A to B in 2 ways, and B to D in 3 ways, there are 6 paths from A to D.

10 down, 5 right. I feel like an idiot now.

Let's try and find out why I can't count this. If I am on A, and I go down 3, and right 4, then I am on B. After I go down 3, I count that same cell as 1 for the right. So I am in column 4, not column 5 (if it is 1-indexed) no?

I see. This is because we could technically go right a bunch from A, then straight down. This makes sense. So if we want to calculate A to D, through B, we could calculate the number of routes from A to B, which is 35. Next we could calculate the number of routes from B to D. To do this I see...

But we want to avoid B for part b, correct? So I would think we want to calculate the number of routes from A to D, and just subtract the routes from A to B?