There are 7 red and 5 blue marbles in a jar. In how many way

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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

OA is D
Last edited by rakeshd347 on Mon Sep 30, 2013 5:18 am, edited 1 time in total.

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by theCodeToGMAT » Sun Sep 29, 2013 7:55 pm
7R, 5B, --> 8 SELECT

1r3b + 2R2B + 3r1B --> lEFT

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

=> 7 * 10 + 21 * 10 + 35 * 5

=> 70 + 210 + 175 = 455

Answer [spoiler]{D}[/spoiler]
R A H U L

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by ganeshrkamath » Mon Sep 30, 2013 12:48 am
rakeshd347 wrote: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

OA coming soon.
To select 8 marbles from 7 red and 5 blue:
6R and 2B => 7C6 * 5C2 = 7*10 = 70
5R and 3B => 7C5 * 5C3 = 21*10 = 210
4R and 4B => 7C4 * 5C4 = 35*5 = 175
Total = 70+210+175 = 455

Choose D

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by GMATGuruNY » Mon Sep 30, 2013 3:07 am
rakeshd347 wrote: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
Good ways = Total ways - Bad ways.

Total ways:
Number of ways to select 8 marbles from 12 options = (12C8) = (12*11*10*9*8*7*6*5)/(8*7*6*5*4*3*2*1) = 495.

Bad way 1: No RED marbles left in the jar
For none of the 7 red marbles to be left in the jar, exactly 1 of the 5 blue marbles must be selected.
Number of ways to choose 1 blue marble from 5 options = 5C1 = 5.

Bad way 2: No BLUE marbles left in the jar
For none of the 5 blue marbles to be left in the jar, exactly 3 of the 7 red marbles must be selected.
Number of ways to choose 3 red marbles from 7 options = 7C3 = (7*6*5)/(3*2*1) = 35.

Good ways:
Total ways - bad ways = 495 - 5 - 35 = 455.

The correct answer is D.
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