10 students, 4 French, others Spanish/German

[Wilder]

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

[StaceyKoprince]

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.

[Wilder]

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.

[shaji]

"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.

[Wilder]

Makes sense - thanks.

[StaceyKoprince]

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.

[Guest]

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.

[StaceyKoprince]

Very nice!

[DaveGill]

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.

[RonPurewal]

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).

[imanemekouar]

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?

[sanidhya510]

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.

[RonPurewal]

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.