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# There are five flavors of icecream: banana, chocolate, lemon

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Intern
Joined: 24 Sep 2009
Posts: 21
There are five flavors of icecream: banana, chocolate, lemon  [#permalink]

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24 Sep 2009, 20:06
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(N/A)

Question Stats:

100% (00:36) correct 0% (00:00) wrong based on 3 sessions

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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
Intern
Joined: 27 Aug 2009
Posts: 35

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25 Sep 2009, 09:58
7
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.
##### General Discussion
Intern
Joined: 24 Sep 2009
Posts: 21

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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

-----
Intern
Joined: 07 Oct 2009
Posts: 16

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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.
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15 Nov 2013, 06:04
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Re: Permutation / combinations   [#permalink] 15 Nov 2013, 06:04
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