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In ti-83/84 notation, prove SUM(SEQ((N nCr K)^2,K,0,N) = (2N) nCr N

or, prove that if you take the nth row of pascals triangle, square each term, then take the sum, the result will be the nth element in row 2n

I cant come up with a method to proove it, anybody know how?

I've asked this twice before a while ago, and the answers I got still don't satisfy so I'll give it a third shot.

You have 2 six sided dice with removable faces. You remove all faces off of both dice and then randomly apply the faces back on again [so you COULD (not necessarily) have the dice looking like 1, 1, 2, 3, 4, 4 and 2, 3, 5, 5, 6, 6 or some other combination.]... What is the probability of rolling a 7?

Thanks!

Here's a few edits/additional information you might find helpful:

Okay, you have TWO six sided dice and you add the numbers rolled on each of them. You cannot simply find one dice combination and calculate the probability from that, the question asks if you randomly apply the faces of each die and have a random roll what is the probability. That's the thing that makes this problem mind boggling though- the sheer number of dice combinations you can have. There's some trick to the problem I know, as the test it appeared on was a no calculator test. (AIME)

If it helps anyone, I looked back through my notebook and I found a few things about this problem.. I did these by complete brute force method in hopes of finding a pattern with combinations or the like, but alas, no luck :P... but one of you may see a pattern. I've done the probabilities for theoretical 2 sided, 3 sided, and 4 sided dice. Doing 5 sided brute force would be hellish, and 6 sided would just defeat the point of finding the trick, not to mention it would take a ridiculous amount of time xD.. so anyways. For a 2 sided die [Btw, all of these start with 0, then go up, so 2 sided die has 0 and 1 on it.], the probabilities are

1/8=0

6/8=1

1/8=2

For a 3 sided die, the probabilities are:

1/15=0

4/15=1

5/15=2

4/15=3

1/15=4

For 4 sided dice, the probabilities are:

1/28=0

4/28=1

5/28=2

8/28=3

5/28=4

4/28=5

1/28=6

Okay, I am having some issues with my IBM thinkpad t43. This is what happened:

Working fine until a week or so ago, at which point the computer randomly shut off. It did not blue screen, or show anything. It instantly black-screened, so some sort of hardware failure. Afterwards, the AC adapter did not charge the battery or the laptop alone--However, the laptop ran fine on battery power. So I figured it was an adapter issue, and accordingly bought a new adapter with the same specs as the other one. The new one does not work either. I took apart the computer and have pretty much determined that there is some issue with the power supply between the AC adapter in and the battery. Any suggestions or pointers as to what I should try troubleshooting would be very much appreciated. thank you! :)

Ok, this is a probability question that one of my friends pondered me and I wasn't able to figure it out after half heartedly working on it for a LONG time. So I was wondering if any of you could enlighten me ^_^... here goes nothing :P

You have 2 six sided dice with removable faces. You remove each face off of both dice and then randomly apply the faces back on again [so you could have the dice looking like 1, 1, 2, 3, 4, 4 and 2, 3, 4, 5, 5, 6, 6 etc. etc.]... What is the probability of rolling a 7?

Thanks!