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# Permutation / combinations

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Permutation / combinations [#permalink]  24 Sep 2009, 20:06
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There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

Why is the answer 35 (7!/3!x4!) and not 5^3 written like (5*5*5)=125?

Thank you
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Re: Permutation / combinations [#permalink]  24 Sep 2009, 22:56
benjiboo wrote:
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

Why is the answer 35 (7!/3!x4!) and not 5^3 written like (5*5*5)=125?

Thank you

It is combination case, isn't it? but I got 25 ways instead of 35, my process to come to this result as follow, not sure where is it went wrong.
B C L S V
3 0 0 0 0 -->C(5,1)= 5 ways
2 1 0 0 0 -->C(5,2)=10 ways
1 1 1 0 0 -->C(5,3)=10 ways
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Re: Permutation / combinations [#permalink]  25 Sep 2009, 07:11
benjiboo wrote:
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

Why is the answer 35 (7!/3!x4!) and not 5^3 written like (5*5*5)=125?

Thank you

The reason you can't use 5x5x5 is that you will have duplicates. For example:
Banana Choc Lemon is essentially the same as Lemon, Choc, Banana and also Choc, Lemon, Banana.
So you do not duplicate it's easier to use the 7C3 formula which gives 35 (7!/3!x4!)
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Re: Permutation / combinations [#permalink]  25 Sep 2009, 09:58
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benjiboo wrote:
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

Why is the answer 35 (7!/3!x4!) and not 5^3 written like (5*5*5)=125?
Thank you

Question 1
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? (flavors cannot be repeated)
5C3 = 5!/(3! x 2!) = 10

Question 2
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? (flavors can be repeated)
5 x 5 x 5 = 125

Question 3
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? (flavors can be repeated but order doesn't matter)
(5+3-1)! / [3! x (5-1)!] = 35

From GMAT point of view, you need to know only first and second questions.

I am sure, 3rd question is not from any reputed GMAT related material.

There are many things, Permutation with repetition, Permutation without repetition, Combination with repetition, Combination without repetition. GMAT doesn't ask all the things. This will become very very complicated.

Still if u want to do research on question number 3, you can do that (but not for your GMAT exam). This question is about "Combination without repetition". So, here we can do repetitions. But point to be noted is that it is a combination problem. So, order doesn't matter.
Normally, when we take 5^3, order matters. So, BBL and BLB are different possibilities. But in 3rd question, BBL & BLB are same only (because order doesn't matter). This way, we need to proceed.
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Re: Permutation / combinations [#permalink]  25 Sep 2009, 15:35
Oops typo in the above. I believe its 5C3 as well. (5!/3!2!) not sure where I got 7 from
But I think TraderAK is right.
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Re: Permutation / combinations [#permalink]  25 Sep 2009, 15:44
yangsta8 wrote:
Oops typo in the above. I believe its 5C3 as well. (5!/3!2!) not sure where I got 7 from
But I think TraderAK is right.

the 7 was right. the 5 is wrong. There are 35 combination.
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Re: Permutation / combinations [#permalink]  25 Sep 2009, 22:34
benjiboo wrote:
yangsta8 wrote:
Oops typo in the above. I believe its 5C3 as well. (5!/3!2!) not sure where I got 7 from
But I think TraderAK is right.

the 7 was right. the 5 is wrong. There are 35 combination.

7C3 is not the solution.

Its just by chance that 7C3 is 35, which is matching with the answer 35 (answer to 3rd question).
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Re: Permutation / combinations [#permalink]  26 Sep 2009, 01:04
Here is from the original text:

Explanation to the problem:

Combination with repetition:

(n+r-1)! / r!(n-1)!

where n is the number of things to choose from, and you choose r of them
(Repetition allowed, order doesn't matter)

(5+3-1)! / 3!(5-1)! =

7! / 3! x 4! =

5040 / 6 x 24 =

35

-----
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Re: Permutation / combinations [#permalink]  26 Sep 2009, 01:12
Hi Benjiboo where is this question from if you don't mind me asking?

I've never come across the formula:
(n+r-1)! / r!(n-1)!
Although that is probably due to my lack of knowledge.

The only two formulas I've come across in OG and Kaplan so far are the two standard formulas of
k!/n!(k-n)! and k!
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Re: Permutation / combinations [#permalink]  29 Sep 2009, 03:26
benjiboo wrote:
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

Why is the answer 35 (7!/3!x4!) and not 5^3 written like (5*5*5)=125?
Thank you

Question 1
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? (flavors cannot be repeated)
5C3 = 5!/(3! x 2!) = 10

Question 2
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? (flavors can be repeated)
5 x 5 x 5 = 125

Question 3
There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be? (flavors can be repeated but order doesn't matter)
(5+3-1)! / [3! x (5-1)!] = 35

From GMAT point of view, you need to know only first and second questions.

I am sure, 3rd question is not from any reputed GMAT related material.

There are many things, Permutation with repetition, Permutation without repetition, Combination with repetition, Combination without repetition. GMAT doesn't ask all the things. This will become very very complicated.

Still if u want to do research on question number 3, you can do that (but not for your GMAT exam). This question is about "Combination without repetition". So, here we can do repetitions. But point to be noted is that it is a combination problem. So, order doesn't matter.
Normally, when we take 5^3, order matters. So, BBL and BLB are different possibilities. But in 3rd question, BBL & BLB are same only (because order doesn't matter). This way, we need to proceed.

Agree with the great explanation given.
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Re: Permutation / combinations [#permalink]  29 Sep 2009, 03:41
what is the solution to the original problem?..
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Re: Permutation / combinations [#permalink]  29 Sep 2009, 20:00
manojgmat wrote:
what is the solution to the original problem?..

In GMAT, if this question comes and its not clarified whether the flavors can be repeated or not, then u should assume that flavors can be repeated.
Answer will be 5*5*5 = 125

However, if there are 5 colored balls and we need to choose 3, and nothing is mentioned whether repetition is allowed or not. In that case, u should assume that balls are unique and each colored ball can not be repeated.
Answer will be 5C3 = 10

I think its clear what is the difference between these two examples. So, original question's answer is 125
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Re: Permutation / combinations [#permalink]  15 Oct 2009, 03:51
The question can be solved with the use of the false coins method.

Now u have three scoops to choose and 5 diff flavours from which to choose from

So u need to put three good coins represented by "O" and 4 false coins represented by "X"

so now we have to arrange these three good coins and 4 false coins

This can be done in 7C3 ways

Now how will this solve the problem

Consider a combination

OOXXOXX

Any "O" left of the first X is the number of scoops of the first flavour = 2 scoops of flavour 1
Any "O" left of second X and right of first X is number of scoops of second flavour = 0 scoops of flavour 2
Similarly any "O" left of third and right of second X is number of scoops of third flavour = 1 scoops of flavour 3
Similarly any "O" left of Fourth and right of third X is number of scoops of fourth flavour = 0 scoops of flavour 4
Now finally any "O" right of fourth X is number of scoops of flavour 5 = 0 scoops of flavour 5 in this case

Similarly we can arrange as
XOXOXOX
Flavour 1 = 0
Flavour 2 = 1
Flavour 3 = 1
Flavour 4 = 1
Flavour 5 = 0

I hope this helps....

Similarly if there were 4 scoops to choose and 6 flavours the answer would be 9C4 as there would be 5 false coins ("X") and 4 good coins ("O") which we have to arrange.
Re: Permutation / combinations   [#permalink] 15 Oct 2009, 03:51
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