From a class consisting of 12 computer science majors, 10 ma

Question

From a class consisting of 12 computer science majors, 10 mathematics majors, and 9 statistics majors, a committee of 4 computer science majors, 4 mathematics majors, and 3 statistics majors is to be formed. How many distinct committees are there?

ulm · Answer

Technically it is
 4C12*4C10*3C9 .
But how to solve it under 2 mins?

Bunuel · Answer

C^4_{12}*C^4_{10}*C^3_{9}=\frac{12!}{8!*4!}*\frac{10!}{6!*4!}*\frac{9!}{6!*3!}=495*210*84 , where  C^4_{12}  is # of selections of 4 computer science majors out of 12,  C^4_{10}  is # of selections of 4 mathematics majors out of 10, and  C^3_{9}  is # of selections of 3 statistics majors out of 9.

petrifiedbutstanding · Answer

Thanks Bunuel!

What is the likelihood of such an absurd calculation on the GMAT.. Any idea?

Bunuel · Answer

You won't be asked to  actually calculate  495*210*84=8,731,800. Though there might be some tricky options:

A. 8,731,600
B. 8,731,700
C. 8,731,800
D. 8,731,900
E. 8,732,800

Now, even not calculating actual value of 495*210*84 we can say that the result must be multiple of 3 (as for example 210 is a multiple of 3). A number is a multiple of 3 if the sum of its digits is a multiple of 3: only option C satisfies this, so it must be a correct answer.

petrifiedbutstanding · Answer

You mean C, right.. Thanks. That's a great take.

KarishmaB · Answer

It is extremely unlikely that the options would be a straight number... Though it is also unlikely that they will directly hand over the answer to you in the above format.

The correct option could be similar to:  \frac{{12*11*10^2*9!^3}}{{8!*6!^2*4!^2*3!}} 
or
 \frac{{12*11*10^2*9^3*8^2*7^2}}{{4!*4!*3!}}  or some such variation...

Rule of thumb: If you need to multiply anything more complex than 2 digit number by another two digit number, there is something you have missed.

aks1985 · Answer

why do you say that the result must be a multiple of 3?

Stoneface · Answer

From a class consisting of 12 computer science majors, 10 mathematics majors, and 9 statistics majors, a committee of 4 computer science majors, 4 mathematics majors, and 3 statistics majors is to be formed. How many distinct committees are there?

Alright, I just started learning combinations tonight, but I have hit a wall. I ran across this problem but cannot get the answer right. Apparently, the answer is 8-some-thousand, but I keep coming up with 16,200. This is how I am working it.

12!/4!(12-4)! * 10!/4!(10-4)! * 9!/3!(9-3!)

(12*...*1/4*...*1*8*...1) * (10*...*1/4*...1*6...*1) * (9*...1/3*2*1*6*...1)

(12*10*9/4*3*2*1) * (10*9*8*7/4*3*2*1) * (9*8*7/3*2*1)

(3*5*3/1*1*1) * (5*3*7/1*1*1) * (3*4*7/1*1*1)

45 * 30 * 12

Answer: 16,200

Any idea where I am messing up?

KarishmaB · Answer

You are trying to figure out what 495*210*84 is. The given options are 
A. 8,731,600
B. 8,731,700
C. 8,731,800
D. 8,731,900
E. 8,732,800

You see that the result will end in 2 zeroes, one from 210 and the other from 5*4 of 495 and 84. All options end in 2 zeroes.

A number is a multiple of 3 if the sum of its digits is a multiple of 3.
You see that 210 is divisible by 3 because 2+1 = 3 is divisible by 3. If 210 is divisible by 3, then it means 3 must be a factor of 495*210*84.
3 is a factor of only 8,731,800 since 8+7+3+1+8 = 27. Hence answer must be 8,731,800.

Sovjet · Answer

Well, I can tell you that your algebra is a little off:

12!/4!8! = (11 x 9 x 5) = 495 (12 cancels 3 and 4, 10 cancels with 2 -> 5)
10!/4!6! = (10 x 3 x 7) = 210 (9 cancels with 3 -> 3, 8 cancels with 2 and 4)
9!/3!6! = (3 x 4 x 7) = 84 (9 cancels with 3 -> 3, 8 cancels with 2 -> 2)

495 x 210 x 84 = 8,731,800

You were cancelling multiple numerator items against the same denominator item (i.e. cancelling both 10 and 8 against the same two - in the second one) as well as forgetting the 11 in the first combination.

As far as I know, that would be the answer... however it seems like I'm a little over what the answer provides. (8,000 you were saying?)

Good way to check your permutations: https://www.mathsisfun.com/combinatorics ... lator.html

fluke · Answer

As sovjet already mentioned, mistake in calculation, not in the application of formula.

C^{12}_{4}*C^{10}_{4}*C^{9}_{3}

495*210*84=8,731,800

**************************************************

Let me simplify one of these for you:
 C^{12}_{4}=\frac{12!}{4!8!}=\frac{12*11*10*9}{4*3*2*1}=\frac{12}{4}*11*\frac{10}{2}*\frac{9}{3}=3*11*5*3=9*5*11=45*11=45*10+45=450+45=495

Sovjet · Answer

Ah, just noticed the and tags. That'll come in handy...

Stoneface · Answer

But how do you get from "12!/4!(12-4)! * 10!/4!(10-4)! * 9!/3!(9-3!)" to "495*210*84"? Were my steps incorrect or did I just totally miscalculator or what? I feel like there is a big elephant in the room and I am not noticing.

Stoneface · Answer

Nix that last post. I went back and did all three fractions the way you showed me and I got the right answer. Now I need to go back and figure out where I originally messed up!

Thanks for your help.

loser · Answer

Thanks for the help.

ranaazad · Answer

Hi bunuel, 
Please let me know why we're not diving this by 3! to avoid repetition.

Thank you for your contribution to this forum.
Regards.

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From a class consisting of 12 computer science majors, 10 ma

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