Register    Login    Search    Rss Feeds

 Page 1 of 1 [ 9 posts ] 



 
Author Message
 Post subject: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Fri Oct 07, 2011 3:54 pm 
Offline
Course Students


Posts: 6
Hi, I have been getting very good at using the line method when doing permutation and combinatoric questions until I came across this relatively easy question (the source is Master the GMAT 2011 Practice Test 4 Problem number 9).

An ice-cream sundae consists of two ice cream scoops, one flavor per scoop, and one topping. How many different types of sundaes can be prepared if four ice-cream flavors and two toppings are available?
A)12
B)14
C)16
D)18
E)20

The correct answe is E) 20 and by manually counting I can get to this answer. For more complex questions, I have been using the line method. So I wanted to attempt to use it for this problem but I am having trouble accounting for repetition.
In this problem you can have 2 scoops of the same flavor, for example vanilla and vanilla.

So far, this is what I came up with
For the flavors (4 *4)/(2*1)
For the toppings 2

I divided by 2*1 because I wanted to eliminate the element of order when choosing flavors.

I thought I already accounted for repetition by multiplying 4*4 in the numberator instead of 4*3, but given the answer (10) this is wrong (with this approach the answer will come up with 16, which is apparently wrong)

Now the question is: What do I do to account for the fact that you can have repeat flavors.

Thanks in Advance


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Thu Oct 13, 2011 11:59 am 
Offline
Students


Posts: 5
This is interesting.

The answer that I though is (4*4*2)/2=16.
Can't figure out how it happens to be 20.


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Wed Nov 23, 2011 1:30 pm 
Offline
ManhattanGMAT Staff


Posts: 2183
Location: Southwest Airlines, seat 21C
Okay on this one I think the approach that makes the most sense is to count how many repeats you have and how many sundaes with no repeats. 4 flavors means 4 possible flavor combinations (4C1) if you have two of the same flavor. Multiply by two toppings to get 8 of course. If we have two different flavors there are 6 combinations (4C2), multiplied by two toppings of course to get 12. Add 8 and 12 to get the 20. Let me know if you still have questions..

_________________
Tim Sanders
Manhattan GMAT Instructor


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Wed Feb 08, 2012 4:24 pm 
Offline
Forum Guests


Posts: 1
mahajanakhil1985 wrote:
This is interesting.

The answer that I though is (4*4*2)/2=16.
Can't figure out how it happens to be 20.


I also got stuck at 16 :(

[irrelevant link deleted]


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Fri Feb 17, 2012 5:53 pm 
Offline
ManhattanGMAT Staff


Posts: 2183
Location: Southwest Airlines, seat 21C
i'm guessing you're a spammer because your username is suspiciously similar to the spammy URL i deleted from your post. if you actually have a legitimate question, see my post above and let me know where you had trouble with the explanation and i'll be glad to help you..

_________________
Tim Sanders
Manhattan GMAT Instructor


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Tue Feb 28, 2012 3:05 pm 
Offline
Course Students


Posts: 36
tim wrote:
Okay on this one I think the approach that makes the most sense is to count how many repeats you have and how many sundaes with no repeats. 4 flavors means 4 possible flavor combinations (4C1) if you have two of the same flavor. Multiply by two toppings to get 8 of course. If we have two different flavors there are 6 combinations (4C2), multiplied by two toppings of course to get 12. Add 8 and 12 to get the 20. Let me know if you still have questions..


what does 4C1/4C2 mean? i still dont get this one. how do you know to use this technique vs just slot method or anagramming?


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Thu Mar 01, 2012 10:01 pm 
Offline
Forum Guests


Posts: 1
I don't get it either


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Fri Mar 02, 2012 1:21 pm 
Offline
Course Students


Posts: 36
My attempt would be this:

4 flavors, in which you choose 2, which can be the same.

4!/(2!)(2!)= 6, but then you have to add back 4 because it could be vanilla/vanilla, chocolate/chocolate, etc.

When you divide by the 2! 2!, you are essentially accounting for 2 things: 1) the two not chosen items are in the same category- the "out" group. and 2) the combination Vanilla/Chocolate is the same as Chocolate/Vanilla.

Now, what you are NOT capturing, is that Vanilla/Vanilla and Chocolate/Chocolate are 2 real possibilities. So you need to add back the 4.

6+4 = 10. then you multiply by 2 to get the combination with toppings.

20

I think this is correct, but will wait for confirmation. Thanks!


Top 
 Post subject: Re: Accounting for repitition in the Line Method Combinatorics
 Post Posted: Sat Mar 03, 2012 11:19 pm 
Offline
ManhattanGMAT Staff


Posts: 1857
Here's how I would do it:

1) Let's say my ice cream flavors are vanilla, chocolate, strawberry, and peppermint: VCSP
--Let's also say that the toppings are butterscotch and cherry: BC

2) I can make four sundaes with two scoops of the same flavor: VV, CC, SS, PP. Each of those four can be topped with either B or C. That yields 8 possibilities.

3) I would then use the slot method to figure out the number of sundaes with two different scoops:
4 possibilities for scoop 1
then
3 possibilities for scoop 2
That yields 12 possibilities.

However, the slot method ALWAYS assumes that order matters. Order does not matter here; we just need two scoops and don't care which is first or second. Because order does not matter, we need to adjust the outcome by dividing by the factorial of the number of slots, or 2!.

Thus, 12/2! = 6 different ways to pick two different scoops of ice cream.

(NOTE: This technique can be useful on other combinatorics problems--slot method with adjustment since order doesn't matter).

4) Finally, each of these 6 combinations of two ice creams can be paired with the two toppings; 6 * 2 = 12 sundaes made with two different scoops.

5) 8 (single type of ice cream) + 12 (two different types of ice cream) = 20 total possibilities

_________________
Jamie Nelson
ManhattanGMAT Instructor


Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 9 posts ] 





Who is online

Users browsing this forum: No registered users and 1 guest

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: