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an idiot proof method for permutation problem?

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jzw Senior | Next Rank: 100 Posts Default Avatar
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an idiot proof method for permutation problem?

Post Fri Feb 24, 2012 12:20 pm
Hi - I'm a PR veteran. I'm hoping someone can explain the PR way of doing this, OR a different virtually idiot proof method - anything OTHER THAN visualizing it. I find that viualizing it is not always a simple thing depending on the problem and I don't want to rely on that. The following is the question:

"A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?"

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ubhanja Junior | Next Rank: 30 Posts Default Avatar
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Post Sat Feb 25, 2012 7:10 am
It can be looked at selecting 3 committee from a group of 4 by 4C3 i.e 4 ways.

Then within each committee there are 2 members ; selecting 1 would be 2C1 i.e 2

so total number of ways 4C3 * 2C1 * 2C1 * 2C1 = 4*2*2*2 = 32

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Post Fri Feb 24, 2012 2:05 pm
jzw wrote:
hey mitch - thanks so much!

so if it was just one group of eight people, it would have been 8*7*6/3*2*1 ?
Yes. The number of committees of 3 that can be chosen from 8 people = (8*7*6)/(3*2*1) = 56.

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GMAT/MBA Expert

Post Fri Feb 24, 2012 1:40 pm
jzw wrote:
Hi - I'm a PR veteran. I'm hoping someone can explain the PR way of doing this, OR a different virtually idiot proof method - anything OTHER THAN visualizing it. I find that viualizing it is not always a simple thing depending on the problem and I don't want to rely on that. The following is the question:

"A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?"
Step 1: Count the number of options for each position.
1st person = 8. (Any of the 8 people.)
2nd person = 6. (Any of the 6 people not on the 1st person's team.)
3rd person = 4. (Any of 4 people on the two teams not yet represented on the committee.)
To combine these options, we multiply:
8*6*4.
Do not calculate the product just yet.

Step 2: Determine whether order matters.
The product above represents the number of ways to ARRANGE the 3 people being chosen.
But since we're choosing a committee, order doesn't matter.
ABC and CAB are the SAME committee.
To account for all of duplicate committees contained in our product above, we must divide by the number of ways to arrange the 3 people being chosen.
The number of ways to arrange 3 elements = 3! = 3*2*1.

Step 3: Calculate the result.
The number of possible committees = (8*6*4)/(3*2*1) = 32.

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If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.
Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.

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jzw Senior | Next Rank: 100 Posts Default Avatar
Joined
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3/16
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760+
Post Fri Feb 24, 2012 1:53 pm
hey mitch - thanks so much!

so if it was just one group of eight people, it would have been 8*7*6/3*2*1 ?

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