Than you in advance.

(Right now, I am doing my best to solve this problem. any help? (Doh))

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- Oct 20th 2008, 06:03 AMNarekshow that ....
Than you in advance.

(Right now, I am doing my best to solve this problem. any help? (Doh)) - Oct 20th 2008, 06:27 AMddt
As to (a), the combinatorial argument:

What has C(2n, 2) to do with bowl of 2n marbles, numbered from 1 to 2n? Then imagine you divide the marbles over two bowls, each containing n marbles.

As to (b), the algebraic argument: write out the formulas...

PS. It is easier for quoting if you'd type out the exercise instead of attaching it as a JPG. - Oct 20th 2008, 06:58 AMNarek
i solved number B) my friend, but i dont have any idea how to solve A)

can you solve it here and explain please? (Rofl)(Rock)

THANK YOU (Nerd) - Oct 20th 2008, 07:07 AMddt
OK. Going from the hint in my previous post:

C(2n, 2) is the number of ways you can pick 2 marbles out of a bowl of 2n marbles.

Now imagine the marbles divided over 2 bowls, n marbles in each. In how many ways can I then pick 2 marbles in total from these 2 bowls?

a) I can pick 2 marbles from bowl 1, in a number of ways

b) I can pick 2 marbles from bowl 2, in b number of ways

c) I can pick 1 marble from each bowl, in c number of ways

The sum a+b+c must be equal to picking 2 marbles from the one bowl we had in the beginning.

Now, what numbers are a, b and c?