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I like the idea of one variable tracking which numbers were picked and the other how many of each was taken.

Not \cdots but \vdots. Right beneath the second line of LaTeX, instead of those three dots.

Thanks so much for this. I was trying to figure out a GF for this problem, but got nowhere.

By the way, I think \vdots might look better.

Okay, so I think it would go like this:

The number of ways of choosing a combination of flavors is , because those are combination with replacement. Then we also multiply by 3, for the sizes. So, there are combinations. From those we need to choose 12 different ones, which we can do in ways.

So, flavors xi and yi are the same, right?

Agnishom wrote:

I'm using this principle to solve the second problem:

a. Generate a list of five matches.

b. Map a cartesian product over each set of five matches.

c. Check if the result is <= 500

Won't you just get 25 ordered pairs?

Hm, is that Haskell?

Hi ElainaVW

Thanks.

Hi NickMeyers

One question - does it matter if J or S was chosen in the last example?

Hi ElainaVW

Could you hide that?

The one Elaina mentioned above, Mathematica.

Hi NickMeyers

Welcome to the forum!

There are 6304 combinations for the new list.

I'd say no, but I was referring to post #1.

Well, you differentiate term by term.

See you and good luck.

Nice work!

I guess we can safely say the answer is 1/4.

(3,4) is correct. (7,16) seems to be the point that maps to (-3,-8).

Good night.

When you get the chance, please tell us what you used to get the mean of the 1's and 0's.

I have to look at it a bit more.

Glad we're finally online at the same time. It's a bit late over there, isn't it?

Have you tried doing it a few more times?