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Combinatronics help needed

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Combinatronics help needed

by ani781 » Thu Sep 04, 2014 6:53 pm
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. We can have three scoops. How many variations will there be?

I seriously need help in these types of questions. Can some expert please suggest me some more similar problems or some guidance. Many thanks in advance.
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by Brent@GMATPrepNow » Thu Sep 04, 2014 7:38 pm
ani781 wrote:There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. We can have three scoops. How many variations will there be?

I seriously need help in these types of questions. Can some expert please suggest me some more similar problems or some guidance. Many thanks in advance.
There are some missing details here.

Does the order of the 3 scoops matter? For example, is chocolate on top, lemon in the middle and banana on the bottom different from banana on top, lemon in the middle and chocolate on the bottom?

Also, can I have more than 1 scoop of the same flavor?

These issues must be addressed before we can answer the question.

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by [email protected] » Thu Sep 04, 2014 8:18 pm
HI ani781,

Brent is correct - we need the exact wording of the question (and answer choices) so that we know what type of math to do.

Here are some possible outcomes (based on the "intent" of the question):

1) Order matters, duplicates are allowed = (5)(5)(5) = 125 options

2) Order matters, duplicates NOT allowed = (5)(4)(3) = 60 options

3) Order does NOT matter, duplicates NOT allowed = 5!/(3!2!) = 10 options

4) Order does NOT matter, duplicates allowed = 5!/(3!2!) + 2[5!/(2!3!)] + 5 = 35 options

You can see that the specific details of the question will dictate how you have to do the math to get the correct answer.

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by ani781 » Thu Sep 04, 2014 11:29 pm
Thanks Brent / Rich.
I Really appreciate your help.
Unfortunately, I don't have the details of this question.
But somehow believe if I get the logic of the 4th case it would be really helpful..

Best Regards.
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by GMATinsight » Fri Sep 05, 2014 10:23 am
ani781 wrote:There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. We can have three scoops. How many variations will there be?

I seriously need help in these types of questions. Can some expert please suggest me some more similar problems or some guidance. Many thanks in advance.
Also if you willing to learn a formula nCr = n!/[r! x (n-r)!]

Then I must tell you that when you have more options and you want lesser without any arrangements) then you may dirctly apply this formula which is very helpful for being efficient at P&C

Here the answer = 5C3 = 5!/(2!3!) = 10
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by [email protected] » Fri Sep 05, 2014 10:27 am
Hi ani781,

If the original question that you posted was written in that exact fashion (from whatever the original source was), then that source should be considered unreliable. There are a lot of substandard GMAT materials floating around the internet, so you have to be wary of any material that does not come from a reputable source. My guess is that this question was meant to be a standard combinatorics question (meaning that the 3 scoops had to be different and the order of the scoops doesn't matter), but maybe it was meant to be something else. The GMAT questions that you'll see on Test Day are specifically worded to remove any confusion or bias on the part of the Test Taker.

As to your question, the 4th "case" that I listed is a far rarer option than the other 3 cases. In that last option, the math treats the situation as a combinatorics question, so order does NOT matter, AND allows for duplicates. This means that there are 3 separate calculations:

First part: All 3 scoops are different.
Second part: 2 scoops are the same and the 3rd is different
Third part: All 3 scoops are the same.

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by Matt@VeritasPrep » Sat Sep 06, 2014 10:47 am
The implication is that the order doesn't matter, but we could use the answer choices to help us.

If all three scoops are the SAME flavor, we have 5 options.

If TWO scoops are the same and the third is different, we have 20 options. (We have FIVE choices for the two identical scoops and FOUR for the other, so 5*4.)

If none of the scoops are the same, we have (5 choose 3) options, or 10 options.

So presuming the order doesn't matter, we have 35 possibilities.

If the order DOES matter, we'd take each case and determine the orders.

If all three scoops are the same, the order is irrelevant, so we still have 5 options.

If two scoops are the same and the third differs, we have 20 options, each of which could be arranged three ways (XXY, XYX, YXX). That gives us 20 * 3 = 60 options.

If all three are different, we have 10 options, each of which can be arranged 6 ways (XYZ, XZY, YXZ, YZX, ZXY, ZYX). This is another 60 options.

Our total here is 60+60+5, or 125.
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