If you're experiencing a roadblock with one of the Manhattan Prep GMAT math strategy guides, help is here!
lkzhang
Students
 
Posts: 2
Joined: Thu Jul 08, 2010 7:34 pm
 

Guide 4 page 185

by lkzhang Mon Nov 15, 2010 3:39 pm

The question is long, the first sentence reads as "Alicia lives in a town..."

I am completely confused on the explanation, especially the sencond part regarding the numerator, why the total blocks reduced to 4 instead of 5? If the total blocks reduced to 4, why the denominator is 3!1!, not 2!2!? How do you decide that the first block is taken for south (as reduced the number of south blocks to 1!), not for east? (which will reduce the number of east blocks to 2?)

Additonly, why Alicia only needs to make 6 decision? As explained in the text that there are only 6 blocks so she needs to make 6 decision. Does she need to make a decision at every corner of the block (except the coners in the last colum or row).

Sorry if my questions look extremely strange, I was just lost. Thanks
jnelson0612
ManhattanGMAT Staff
 
Posts: 2664
Joined: Fri Feb 05, 2010 10:57 am
 

Re: Guide 4 page 185

by jnelson0612 Wed Nov 17, 2010 4:09 pm

lkzhang, let me try to answer these and please write back if you need more clarification.

Why does Alicia need to make 6 decisions? Because she needs to walk 6 blocks home, with 3 east and 3 south, and for each block go either east or south. You are right that at each corner of a block she needs to make a decision as to where she is going to go; thus, she will make six decisions. 6!/3!3! will give us the total number of possible routes.

The problem asks us the probability of Alicia walking south the first two blocks. Once we assume she is going to walk south for the first two blocks, we know that her eventual route will look like this:
SS_ _ _ _.
Thus, we have four blocks remaining, but I know that of her six total blocks she had to walk three east and three south. We have already removed two of the south blocks (they are in places 1 and 2), so we have three east and one south block remaining.

Thus, her remaining number of combinations is:
4! (number of blocks remaining to be walked)
3! (# of east blocks remaing) 1! (# of south blocks remaining)

I hope this helps!

Thank you,
Jamie Nelson
ManhattanGMAT Instructor
saurabhbanerjeeiimk
Forum Guests
 
Posts: 27
Joined: Sat Oct 01, 2011 5:23 pm
 

Re: Guide 4 page 185

by saurabhbanerjeeiimk Fri Nov 04, 2011 6:31 am

A little confused!

The question asks about the probability for the first two south blocks. Why are we calculating for the remaining 3?
StaceyKoprince
ManhattanGMAT Staff
 
Posts: 9360
Joined: Wed Oct 19, 2005 9:05 am
Location: Montreal
 

Re: Guide 4 page 185

by StaceyKoprince Sun Nov 13, 2011 11:17 pm

Because there are multiple ways in which she can walk all 6 blocks *after* walking south for the first 2 blocks. Probability is a measure of the desired outcomes / all possible outcomes, so we have to calculate all of the desired outcomes and all of the possible outcomes in order to solve.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep