Recent content by CountNumberla

The book we use is VERY vague, with only a few examples, and we've only gone as far as the equation I've shown you. I don't understand why you can't help me with ONE answer, it's not like I haven't stressed myself trying to find the solution! I mean you rather I never figure it out?

some permutations yes, but not that formula. However, I know a similar one. so back to the original problem, 10 red and 6 blue, in how many ways can you pick 2 of the same color? is it 16! / 10! 6! = 56 I'm not sure how to compute picking different colors though... And are the others...

So it's safe to say, 6! - (5! x 2) = 480 Hmmmm 2 red marbles = 2 x 10 = 20 2 blue marbles = 2 x 6 = 12 thanks

I got it thanks! 2000(1 + .24/12)^36 = $4079.77 25[(1 + .24/12)^36 - 1] / (.24/12) = $1299.86 Therefore 4079.77 - 1299.86 = $2779.92 correct?

I understand, if you see my other post, I AM doing the work myself. Here's the equation I used: 2000(1 + .24/12)^36 = X [(1 + .24/12)^36 - 1] / (.24/12) Im not sure what other equations to use for the 2nd problem. thanks!

Hi again! Bob's credit card balance is $2000. Credit company charges 24% interest compounded monthly. 1) Amortize monthly payments to pay off in 3 years I'm not sure how to write the equations on the board, but I get $78.47 2) If Bob only pays $25 monthly minimum, what will his debt...

Thanks LCKurtz, OK, here's what I got:

PROBLEM 1 7 people sit in a row, and 2 of them want to sit together, in how many ways can they be arranged? ATTEMPTED SOLUTION: 6! x 2 = 1440 ------------------------------------------------------------------------------------------------------- PROBLEM 2 7 people sit in a row, and 2 of them...