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by gmat6087 » Wed Oct 24, 2012 4:44 am
5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?
A. 460
B. 490
C. 493
D. 455
E. 445

OAD

My question is can we solve it as given below

7c4*5c4 + 7c5*5c3 + 7c6*5c2

Please respond
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by anuprajan5 » Wed Oct 24, 2012 5:57 am
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]
Regards
Anup

The only lines that matter - are the ones you make!

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by AIM TO CRACK GMAT » Wed Oct 24, 2012 6:09 am
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...
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by anuprajan5 » Wed Oct 24, 2012 6:19 am
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.
Regards
Anup

The only lines that matter - are the ones you make!

https://www.youtube.com/watch?v=kk4sZcG ... ata_player
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