Accounting for repitition in the Line Method Combinatorics

[a.sarwari]

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

[mahajanakhil1985]

This is interesting.

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

[tim]

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..

[aka-accounting2]

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]

[tim]

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..

[NMencia09]

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?

[vritzka]

I don't get it either

[NMencia09]

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!

[jnelson0612]

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