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# M26-22

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Math Expert
Joined: 02 Sep 2009
Posts: 46205

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16 Sep 2014, 01:25
1
1
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Difficulty:

75% (hard)

Question Stats:

57% (02:10) correct 43% (03:04) wrong based on 35 sessions

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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 remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445

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Math Expert
Joined: 02 Sep 2009
Posts: 46205

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16 Sep 2014, 01:25
Official Solution:

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 remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445

Total ways to select 8 marbles out of $$7+5=12$$ is $$C^8_{12}$$;

Ways to select 8 marbles so that zero red marbles is left in the jar is $$C^7_7*C^1_5$$;

Ways to select 8 marbles so that zero blue marbles is left in the jar is $$C^5_5*C^3_7$$;

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

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Joined: 19 Dec 2015
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GMAT 1: 720 Q50 V38
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09 Mar 2016, 21:05
1
Bunuel,

This is a duplicate question. Please refer to M13-37.
Math Expert
Joined: 02 Sep 2009
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09 Mar 2016, 22:49
FacelessMan wrote:
Bunuel,

This is a duplicate question. Please refer to M13-37.

Thank you. Removed M13-37 from the data base.
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Joined: 17 Sep 2014
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Concentration: Operations, Strategy
GMAT 1: 710 Q49 V38
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11 Oct 2016, 20:54
Bunuel wrote:
Official Solution:

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 remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445

Total ways to select 8 marbles out of $$7+5=12$$ is $$C^8_{12}$$;

Ways to select 8 marbles so that zero red marbles is left in the jar is $$C^7_7*C^1_5$$;

Ways to select 8 marbles so that zero blue marbles is left in the jar is $$C^5_5*C^3_7$$;

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

The question says at least one red marble AND at least one blue marble.
But you add the possibilities (5+35). Shouldn't it be multiplied ?
Math Expert
Joined: 02 Sep 2009
Posts: 46205

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12 Oct 2016, 00:09
avdgmat4777 wrote:
Bunuel wrote:
Official Solution:

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 remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445

Total ways to select 8 marbles out of $$7+5=12$$ is $$C^8_{12}$$;

Ways to select 8 marbles so that zero red marbles is left in the jar is $$C^7_7*C^1_5$$;

Ways to select 8 marbles so that zero blue marbles is left in the jar is $$C^5_5*C^3_7$$;

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

The question says at least one red marble AND at least one blue marble.
But you add the possibilities (5+35). Shouldn't it be multiplied ?

There are 2 separate cases when there can be 0 red or 0 blue left:
If we pick 7 red (out of 7) and 1 blue.
If we pick 5 blue (out of 5) and 3 red.

These cases are separate, they cannot happen together so we add.
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Joined: 07 Oct 2015
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30 Nov 2016, 02:57
Total ways of selecting 8 marbles=495
Marbles that remain in the jar= 1R and 3B or 3R and 1B= 35/2 *2= 35. In this case shouldn't the answer be 460?
Am I missing something?
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06 Dec 2016, 23:04
I took the long way

Applicable Cases: 4B and 4 Red, 3 Blue and 5 Red, 2 Blue and 6 Red - Remaining two cases with 1B and 7 Red and 5 B and 3 Red Discarded

175 + 210+70 =455
Intern
Joined: 11 Oct 2017
Posts: 11

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13 Jan 2018, 11:41
Bunuel
Are the same colored marbles identical?
I didn't understand how there would be 5 ways to choose a blue marble if they are all identical? I thought there should be 1 way only if the red marbles are identical to each other and blue identical to the other blue.

thank you
Intern
Joined: 03 Sep 2017
Posts: 15

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23 May 2018, 14:54
Another approach:

The selecting/de-arranging method (without C formulas).

Total ways to choose 12 balls.
12*11*10*9*8*7*6*5 / 8! -> 495
the divided by 8! is because we are ''de-arranging'' since the order doesn't matter.

Ways to choose all red:
RRRRRRRB
By formulas would be 7*6*5*4*3*2*1*5 / 7!*1! = 5
7! and 1! same de-arrange concept.

Ways to choose all blue
BBBBBRRR = 5*4*3*2*1*7*6*5 / 5!*3! = 35
5! and 3! same de-arrange concept.

495-35-5 = 455.
Re: M26-22   [#permalink] 23 May 2018, 14:54
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# M26-22

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