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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
