This is a question which I got wrong in GMAT Prep1. Can someone please help me understand this?

the correct answer for the above is 6. Thanks in advance for your replies

2 chairs from 5, 5C2 = 10. The total possibilities are 150, so the possibilities for tables = 150/10 =15. Let N be the number of tables. So NC2 = 15. We know that 6C2 = 15. So N = 6.

-Raj.

Question for Raj:

For 5C2 = 10 => Can this be calculated as a formula or is it simply 5!/2!3!?

How did you calculate the possibilities for tables (150/10 =15). What is the rule here?

Thanks

Emma,

yes, I used 5C2, formula for choosing 2 out of 5 chairs. Alternatively,

the first chair can be chosen in 5 ways, the 2nd in 4 ways, so 5*4 = 20 ways but these ways would include duplicates like Chair1Chair2 & Chair2Chair1, hence divide by 2 => 20/2 = 10.

For your question about 150/10, it is given that the total number of possibilities that the decorator has with tables and chairs is 150. So (possibilities of chairs)*(possibilities of tables) = 150

Hope that helps,
-Raj.

yes, and yes.
the formula is precisely the fraction you've provided.




not sure which part of the process you're asking about, so let's look at the whole thing.

the choices of tables and chairs are INDEPENDENT, so you MULTIPLY (number of ways of choosing tables) x (number of ways of choosing chairs) to get the total number of ways of selecting everything.
therefore, 150 is the product of the number of ways of choosing chairs (10) and the number of ways of choosing tables (unknown).
so the number of ways of choosing tables is 150/10 = 15.
because you're selecting 2 tables from an unknown number of tables (let's call it "n"), you need (n!) / (n - 2)!2! to equal 15.
two ways to find this:
(a) reduce the fraction to n(n - 1)/2 = 15, turn it into a standard quadratic, and solve for n;
(b) take the answer choices, plug them in, and see which one gives you 15. as an extra bonus, you know that the answer is 10 if n = 5 (that's the calculation for the chairs), so you know you're looking for a number that's bigger than 5, but not by too much. therefore, 6 - which turns out to be the correct answer - is a natural first choice.
i like (b) better, but (a) is just as good if you're lightning at quadratics.

Ron,

I apologize but can you please show me how to reduce the factoral in step a? I just solved this by testing 6 (as you discussed in step b).

thanks,

Imraan

Here's the factorial manipulation, Imraan.

Write n! as a product, according to the definition of factorials:
n!/(n-2)!2! = (n)(n-1)(n-2)(n-3)...(2)(1)/(n-2)!2!

Recognize that the factorial (n-2)! is embedded in the numerator, and thus cancels:
(n)(n-1)(n-2)(n-3)...(2)(1)/(n-2)!2! = (n)(n-1)(n-2)!/(n-2)!2! = (n)(n-1)/2!

2! is just 2*1 = 2, so we end up with (n)(n-1)/2.

_________________
Emily Sledge
Instructor
ManhattanGMAT

thanks Emily!

glad it helped