If you're experiencing a roadblock with one of the Manhattan Prep GRE math strategy guides, help is here!
manigraja100
Students
 
Posts: 14
Joined: Sun Mar 03, 2013 12:00 am
 

Word problems.Pg 169.#1.

by manigraja100 Tue Jun 25, 2013 1:48 am

Q- Joe will pick 3 friends to join him on a road trip.His friend group consist of 4 musicians and 3 poets.In how many different ways can joe select his 3 traveling companions so that he has atleast one musicians and atleast one poet among them ?
A-Joe has two ways: “two musician and one poet” or “two poet and one musician”.
I’m using slot method.
For the 1st case 4*3*3 =36 and the 2nd case 3*2*4 =24.So,36+24=60 ways ? But the answer is wrong.Where am i going wrong ?
tommywallach
Manhattan Prep Staff
 
Posts: 1917
Joined: Thu Mar 31, 2011 11:18 am
 

Re: Word problems.Pg 169.#1.

by tommywallach Wed Jun 26, 2013 6:18 pm

Hey Manigraja,

You're forgetting that order doesn't matter. Whenever order doesn't matter (in the slots method), you divide by the # of slots factorial wherever order doesn't matter.

2 poets + 1 musician = (4 * 3 * 2) / 2! = 12
2 musicians + 1 poet = (4 * 3 * 3) / 2! = 18

Keep in mind, we are dividing by 2! because order doesn't matter within the poets (when there are 2) or within the musicians (when there are 2). To see this logically:

Imagine the Musicians and Poets are named like this:

M1, M2, M3, M4
P1, P2, P3

The ways to get 2 poets and 1 musician could be counted out like so:

M1 - P1 - P2
M1 - P2 - P1
etc.

But these two groups are the SAME group of people! So for any given combination of three people, there will always be two ways to write the same group. This is why we divide by 2 (or 2!, technically, because if there were three poets and one musicians, we would be dividing by 3!).

Hope that helps!

-t

P.S. I do prefer the slots method to the equations, just FYI.
manigraja100
Students
 
Posts: 14
Joined: Sun Mar 03, 2013 12:00 am
 

Re: Word problems.Pg 169.#1.

by manigraja100 Fri Jun 28, 2013 11:34 am

Hi Tommy,
Thanks for your explanation.
P.S. i'm following slot method after you . it is flexible.
tommywallach
Manhattan Prep Staff
 
Posts: 1917
Joined: Thu Mar 31, 2011 11:18 am
 

Re: Word problems.Pg 169.#1.

by tommywallach Sat Jun 29, 2013 11:35 am

Agreed! : )

-t
harsha
Course Students
 
Posts: 2
Joined: Sun May 12, 2013 10:48 am
 

Re: Word problems.Pg 169.#1.

by harsha Mon Jul 08, 2013 12:44 am

Hi there,

I was trying out an alternate second approach for the same problem where the nCr formula is directly used, but I'm going wrong somewhere. I appreciate if anyone can point out what I'm missing.

So the way I'm looking at it is -
Slot 1-Slot 2-Slot 3
Slot 1 - The first (at least one) musician can be picked in 4C1 ways (1 musician to be selected from 4 musicians).

Slot 2 - The first (at least one) poet can be picked in 3C1 ways (1 poet to be selected from 3 poets).

Slot 3 - Here either a musician or a poet needs to be chosen from 5 remaining musicians + poets. So this will be 5C1 way.

So the total number ways would be 4C1*3C1*5C1 = 60 ways.

In Tommy's approach above, I agree that it would be the sum of MMP (4C2*3=18) and PPM (3C2*4=12) which is 30.

Thanks in advance.

-H
michael.k.bilow
Manhattan Prep Staff
 
Posts: 8
Joined: Wed Sep 28, 2011 2:49 am
 

Re: Word problems.Pg 169.#1.

by michael.k.bilow Mon Jul 08, 2013 5:31 pm

Hi Harsha,

There is a very subtle error in your reasoning, and it has to do with the fact that the order of selection doesn't matter.

For the most part, you're correct. You can select one musician from 4, and one poet from 3, and one more person from the remaining 5.

Let's call the musicians M1...M4 and the poets P1...P3. Consider the following two ways to choose a group.

M1, then P2, then M4

versus

M4, then P2, then M1.

Both of these selections conform to your rule--pick a musician, then pick a poet, then pick one of the remaining guys. But the groups that you end up with are identical (M4,P2,M1). The question, then, is by how much? You're going to end up either with two musicians or two poets (but not both), and they could come in any order. Therefore, you need to divide your answer 60 by 2! to account for the different possible orderings of the musician or poet.
harsha
Course Students
 
Posts: 2
Joined: Sun May 12, 2013 10:48 am
 

Re: Word problems.Pg 169.#1.

by harsha Tue Jul 09, 2013 7:11 pm

Hi Mike,

Thanks a lot for the response. You are right, I had assumed that there are no repetitions in the number of combinations got from 4C1*3C1*5C1. I had wrongly imagined that after the one Musician and one Poet were chosen from 4C1 and 3C1 ways respectively, they could no longer still be in the remaining 5C1 combinations. In reality 4C1 and 3C1 give the _total_ number of ways, so there is no guarantee that those chosen in 4C1 and 3C1 ways will not reappear in that 5C1 count (i.e., they are not independent events).

I was looking at this problem as one of the digits problems like where we count the number of possible 3 digit numbers. The difference between those digit problems and a problem like poets/musicians is that combinations in the former case are independent while they not independent in the latter case. And this is why I missed handling the repeated objects. Like you pointed out, the number of potential repeated objects in per-combination for this problem is 2 where it could be MM or PP.

Just to complete the analysis, I see two potential approaches to counting combinations (not permutations) in general –

1. Count the total number of combinations and take care of the repetitions by dividing by the number of possible repetitions per-combination. In this example it would be (4C1*3C1*5C1)/2! = 30.
2. Count the exact number of combinations. In this example it would be the sum of number of combinations for MMP (4C2*3=18) and PPM (3C2*4=12) which is 30.

This would not apply to permutations since both AB and BA are valid permutations. This is also loosely analogous to directly finding P(E) or finding 1-P(~E) in order to find P(E).

Thanks again.

-H
tommywallach
Manhattan Prep Staff
 
Posts: 1917
Joined: Thu Mar 31, 2011 11:18 am
 

Re: Word problems.Pg 169.#1.

by tommywallach Thu Jul 11, 2013 9:35 pm

Hey Harsha,

I also want to add, I strongly advise you to switch over to the slots method, rather than the equation methods. The slots method is more logical than equations, and thus more flexible. Also, the error that you made is far less likely to get made on the slots method, because it's part of the steps.

Step 1: Find number of slots (Decision points)
Step 2: Fill in each slot with number of choices at that decision point
Step 3: Figure out if Order matters or not (or a combination of the two); if it helps: if choc-vanill is not the same as vanil-choc, order matters
Step 4: If Order matters, just multiply all slots; If order doesn't matter, multiply all slots and divide by the # of slots factorial wherever order doesn't matter

Good luck!

-t