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# M26-22

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:25
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Difficulty:

85% (hard)

Question Stats:

55% (01:57) correct 45% (02:02) wrong based on 192 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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Joined: 02 Sep 2009
Posts: 49544

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16 Sep 2014, 01:25
1
7
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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09 Mar 2016, 21:05
1
Bunuel,

This is a duplicate question. Please refer to M13-37.
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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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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 ?
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Joined: 02 Sep 2009
Posts: 49544

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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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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
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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
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Joined: 03 Sep 2017
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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.
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16 Aug 2018, 10:55
The question prompt states:"at least one red marble AND at least one blue marble." Yet, the answer seems to consider one red marble OR one blue marble.
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Joined: 26 Jun 2018
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16 Aug 2018, 23:00
3ksnikhil wrote:
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?

2R and 2B is what your'e missing.
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18 Aug 2018, 03:27
Xin Cho wrote:
The question prompt states:"at least one red marble AND at least one blue marble." Yet, the answer seems to consider one red marble OR one blue marble.

Because when you take out the possibility of there being either zero Reds or zero Blues from all available options, it gives you all the other ways (in numbers) where after taking out 8 marbles, there is at least 1 Red and at least 1 Blue left in the jar.
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18 Aug 2018, 03:27
vitorcbarbieri wrote:
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.

This is the same thing as the C formula.
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If you can dream it, you can do it.

Kudos are appreciated.

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Joined: 12 Jan 2017
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11 Sep 2018, 12:48
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 $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

Just to explain this a bit more:

Problem says there are 12 marbles total, 7 red 5 blue.
Problem wants to select 8, so that there are at least 1 red and 1 blue not selected.

When you see "at least" (or "at most") this is a sign that you want to consider doing P(A) - P(not A) to get your answer.
So consider how many ways there are to select 8 marbles out of 12, with no restrictions. This is 12C8, or 12!/4!8! = 495.

Then think about what the problem wants. What is the opposite of the situation in which you have at least one red and at least one blue? The opposite is you have no red and no blue. That is the opposite of at least 1 red and at least 1 blue.

With that in mind, think about how you would get into those situations... and note that these are two separate (OR) situations.
To have zero red left, you need to pick 7 red and 1 blue.
To have zero blue left, you need to pick 5 blue and 3 red.

Convert these both to math...
Zero red = 7C7*5C1 = 7!/7! * 5!/1!4! = 5
Zero blue = 5C5 * 7C3 = 5!/5! * 7!/3!4! =35
Add the two together and you get 40.

Now remember P(A)-P(not A) = answer.
Do the math:
495 - 40 = 455

M26-22 &nbs [#permalink] 11 Sep 2018, 12:48
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