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M26-22

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M26-22 [#permalink]

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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
[Reveal] Spoiler: OA

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Re M26-22 [#permalink]

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New post 16 Sep 2014, 00:25
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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 \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D
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Re: M26-22 [#permalink]

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New post 09 Mar 2016, 20:05
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Bunuel,

This is a duplicate question. Please refer to M13-37.
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Re: M26-22 [#permalink]

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New post 09 Mar 2016, 21:49
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Re: M26-22 [#permalink]

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New post 11 Oct 2016, 19:54
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 \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D


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 ?
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Re: M26-22 [#permalink]

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New post 11 Oct 2016, 23:09
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 \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: M26-22 [#permalink]

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New post 30 Nov 2016, 01:57
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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Re: M26-22 [#permalink]

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New post 06 Dec 2016, 22:04
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
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Re: M26-22 [#permalink]

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New post 13 Jan 2018, 10:41
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 :)
Re: M26-22   [#permalink] 13 Jan 2018, 10:41
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