Lets play marbels

Hi,

Your method is a bit lengthy. How I attacked this was to turn this around.

I figured out total combinations - 12C8 = 495

The combinations where there was no red marble or no blue marble

7C7*5C1+5C5*7C3 = 40

[spoiler]Answer is 495-40 = 455. Hence D[/spoiler]

anuprajan5 wrote:Hi,

Your method is a bit lengthy. How I attacked this was to turn this around.

I figured out total combinations - 12C8 ??= 495

The combinations where there was no red marble or no blue marble

7C7*5C1+5C5*7C3 = 40 ?????????

[spoiler]Answer is 495-40 = 455. Hence D[/spoiler]

Anup... its a very nice method... however.. ill b glad if u cud jus xpand certain thins as m a bit slow at undastandin...

I apologize.

7C7*5C1+5C5*7C3 = 40

All red selected* remaining blue or All blue selectd*remaining red. Basically what I am trying to do is remove the combinations where all the red or all the blue are selected such that the jar contains 0 blue or 0 red.

Since the condition is that it needs atleast 1 blue and 1 red, if you remove these combinations, all other combinations will have atleast 1 red and 1 blue marble in the jar.

I hope that helps.