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Wilder
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Post subject: 10 students, 4 French, others Spanish/German  Posted: Tue Jan 29, 2008 12:14 am |
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Question reads:
10 person group, with 4 French and the others either Spanish or German. Must choose a 3 person committee but one must be French. How many committees are possible?
Answer is 100, but I just can't seem to get there. Please advise.
Thanks - Wilder
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StaceyKoprince
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Post subject: Posted: Tue Jan 29, 2008 6:31 pm |
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| ManhattanGMAT Staff |
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Posts: 6064
Location: San Francisco
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Please make sure to post the entire text of the question, exactly, including answer choices. The specific wording can make a very big difference. In addition, part of your strategy is to look at the answer choices and see what they're like before you choose a particular strategy to approach that question!
Please also make sure to follow protocol: your subject heading for your post should be the first 5-8 words in the question.
_________________
Stacey Koprince
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Director of Online Community
ManhattanGMAT
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Wilder
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Post subject: In a certain group of 10 members, 4 members teach only Fr... Posted: Tue Jan 29, 2008 9:00 pm |
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In a certain group of 10 members, 4 members teach only French and the rest teach only Spanish or German. If the group is to choose a 3-person committee, which must have at least one member who teaches French, how many different committees can be chosen?
A - 40
B - 50
C - 64
D - 80
E - 100
Answer is E.
Thanks.
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shaji
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Post subject: STACEY IS SO RIGHT!!! Posted: Thu Jan 31, 2008 12:27 am |
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"at least" is the crucial word missed outin the earlier version.
The quick fix is the reverse gear!!!
Total # of committees=120(10!/3!/7!) less # of committees with no French=20(6*5*4/6)=100.
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Wilder
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Post subject: Thanks Posted: Thu Jan 31, 2008 5:31 pm |
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StaceyKoprince
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Post subject: Posted: Fri Feb 01, 2008 1:54 am |
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Thanks, Shaji! And, yes, those two little words make a critical difference! As Shaji mentions, if you want "at least one" then that means 1 would work, 2 would work, or 3 would work. Only 0 would not work. It's a lot easier to calculate one option (the 0 scenario) than it is to calculate three different options (1 + 2 + 3)! So do what Shaji did: find the total # and subtract the # that have 0 French and you're done.
_________________
Stacey Koprince
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Director of Online Community
ManhattanGMAT
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Guest
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Post subject: Posted: Wed Feb 20, 2008 7:01 pm |
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You can also use some probability to solve this problem.
The total number of committees has already been discussed... (10!/3!7!) = 120.
Then, since there are 6 non-French teachers, you can calculate the probability that three non-French teachers will be selected in a row... (6/10) x (5/9) x (4/8) = 1/6.
So, 1/6 of the 120 arrangements - or 20 - contain no French teachers. If 20 don't, then the other 100 arrangements do contain French teachers.
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StaceyKoprince
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Post subject: Posted: Fri Feb 22, 2008 1:13 am |
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Posts: 6064
Location: San Francisco
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Very nice!
_________________
Stacey Koprince
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Director of Online Community
ManhattanGMAT
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DaveGill
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Post subject: Posted: Wed Jun 04, 2008 10:54 pm |
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Can some one explain why this is wrong
To ensure atleast one french let us first select one french person and then select 2 more members from all the remaining members.
No. of ways 1 French can be chosen from a group of 4 = 4C1 = 4
No of ways 2 other members can be chosen from 9 remaining members = 9C2 = 36.
Total numbers of ways = 4 x 36 = 144.
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RonPurewal
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Post subject: Posted: Thu Jun 05, 2008 6:59 am |
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DaveGill wrote: Can some one explain why this is wrong
To ensure atleast one french let us first select one french person and then select 2 more members from all the remaining members.
No. of ways 1 French can be chosen from a group of 4 = 4C1 = 4
No of ways 2 other members can be chosen from 9 remaining members = 9C2 = 36.
Total numbers of ways = 4 x 36 = 144.
this doesn't work because, if a group contains more than one french teacher, then that group gets counted more than once.
here's why:
let's say that jacques and jean-baptiste are 2 of the french teachers, and mr. x is a third person.
then your method will count the following situations separately:
1) jacques is selected as the 1 out of 4, and jean-baptiste and mr. x are selected as the 2 out of 9;
2) jean-baptiste is selected as the 1 out of 4, and jacques and mr. x are selected as the 2 out of 9.
but those are the same group.
not good.
you could always salvage this method by subtracting out the # of groups that get double-counted in that way, but that's a lot of effort (not to mention that it requires a masterful intuition for combinatorics).
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imanemekouar
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Post subject: Re: 10 students, 4 French, others Spanish/German Posted: Tue Jan 19, 2010 12:43 am |
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Please can somebody helps.
I have a big problem with probability. I don't know should a use factorial or,when should a use probability . Can you please explain that?
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sanidhya510
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Post subject: Re: 10 students, 4 French, others Spanish/German Posted: Mon Feb 08, 2010 8:38 am |
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dave if thats the you have to approach yhen the "combination" could be used to solve this -
6C2*4C1 (for one french and two others) +
6C1*4C2 (for 2 french) +
4C3 (all three french)
Add all three.
but i personally prefer the all - none approach to tackle these "at least one" problems.
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RonPurewal
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Post subject: Re: 10 students, 4 French, others Spanish/German Posted: Mon Mar 08, 2010 7:44 am |
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imanemekouar wrote: Please can somebody helps.
I have a big problem with probability. I don't know should a use factorial or,when should a use probability . Can you please explain that? the difference between combinatorics (counting) problems and probability problems is extremely well demarcated in the text of the problems. if the problem is asking you for a probability, then you will ALWAYS see the word "probability" in the question stem. in more informal speech there are synonyms for "probability", such as "chance", but on a test as formal as the gmat it will always say "probability" explicitly. if the problem involves combinatorics / counting, then you will see the words "how many" somewhere in the problem. how many such committees are possible?
in how many ways can the cards be chosen?etc.
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