If a quality control check is made by inspecting a sample of 2 light bulbs from a box of 12 light bulbs, how many different samples can be chosen?

a) 6
b) 24
c) 36
d) 66
e) 72

the answer is D

the question is pretty easy and i know the answer is 12 choose 2, but what i'm struggling with is how to do this without a calculator and fast!

i'm also struggling with doing factorials fast without a calculator...

any tips?

bump

i'm going to assume that you can expand the expression "12 choose 2", since you mention it so casually.

this expression expands to (12!) / (10! 2!).

from what you're typing, it appears that you may actually be trying to evaluate this expression by multiplying out the factorials (!!!!) and then performing some sort of long division on the resulting numbers (!!!!).
this is really not the way to go.

instead, you should realize the following major takeaway:
larger factorials will always "swallow" smaller factorials.

here's what i mean: factorials consist of products of ALL positive integers up to and including the given integer. therefore, all of the integers that appear in some smaller factorial will also appear as part of a larger factorial.

so, in the quotient above, all of the numbers from 1 through 10 (from 10!) will also appear in 12!. you can thus cancel those numbers, leaving only the 11 and 12 remaining from 12!.
therefore, the fraction equals (11 x 12) / (1 x 2), or 11 x 6 = 66.

--

if you find yourself actually having to multiply out a factorial that is larger than, say, 5!, then you are probably missing out on LOTS of cancellation.
the gmat is not a test of "grinding arithmetic".

however, if you DON'T KNOW the shortcuts, then, by all means, you should try to grind out the computations. anything is better than sitting there staring at the problem.

thanks RonPurewal, that helps alot...

since we are discussing shortcuts....i came across another problem that went something like..." what is the sum of all integers from 132 to 531 inclusive?"

how do we go about solving this?

i made a new thread about this: click here

from now on, if you have a new question that is completely unrelated to the matter at hand, then you should create a new thread for that question.
thanks.

n!
r!(n-r)!

12!
2!(10)!

12*11(10)!/2(10)!
2*11 = 66

Answer is C

Correct madhan.

_________________
Jamie Nelson
ManhattanGMAT Instructor

I never learned the combination/permutation formulas. In this problem, I just answer like this:

There are 12 choices for the first bulb pulled, and then 11 choices for any second bulb pulled. So there are 12x11=132 possible pairs pulled from the box of 12.

But half of those pairs are the same 2 bulbs in different order (bulb 4 with bulb 2 is the same as bulb 2 with bulb 4). And order doesn't matter in this question.

So divide total outcomes by duplicate outcomes: 132/2 = 66

;)

this is good.

in fact, the vast majority of gmat combinatorics problems actually tend to be simpler WITHOUT the combination and permutation formulas.
for more on this, see the DECEMBER 3, 2009, recording here:
http://www.manhattangmat.com/thursdays-with-ron.cfm