There are five flavors of icecream: banana, chocolate, lemon : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 08:24

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 24 Sep 2009
Posts: 23
Followers: 5

Kudos [?]: 148 [0], given: 2

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

### Show Tags

24 Sep 2009, 19:06
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:36) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

There are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will there be?

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

Thank you
Intern
Joined: 13 Sep 2009
Posts: 16
Followers: 0

Kudos [?]: 4 [0], given: 0

### Show Tags

24 Sep 2009, 21:56
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
Senior Manager
Joined: 31 Aug 2009
Posts: 419
Location: Sydney, Australia
Followers: 8

Kudos [?]: 276 [0], given: 20

### Show Tags

25 Sep 2009, 06:11
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!)
Intern
Joined: 27 Aug 2009
Posts: 46
Followers: 4

Kudos [?]: 97 [6] , given: 1

### Show Tags

25 Sep 2009, 08:58
6
KUDOS
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.
_________________

Salaries are low in recession. So, working for kudos now.

Senior Manager
Joined: 31 Aug 2009
Posts: 419
Location: Sydney, Australia
Followers: 8

Kudos [?]: 276 [0], given: 20

### Show Tags

25 Sep 2009, 14:35
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.
Intern
Joined: 24 Sep 2009
Posts: 23
Followers: 5

Kudos [?]: 148 [0], given: 2

### Show Tags

25 Sep 2009, 14:44
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.
Intern
Joined: 27 Aug 2009
Posts: 46
Followers: 4

Kudos [?]: 97 [0], given: 1

### Show Tags

25 Sep 2009, 21:34
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).
_________________

Salaries are low in recession. So, working for kudos now.

Intern
Joined: 24 Sep 2009
Posts: 23
Followers: 5

Kudos [?]: 148 [0], given: 2

### Show Tags

26 Sep 2009, 00:04
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

-----
Senior Manager
Joined: 31 Aug 2009
Posts: 419
Location: Sydney, Australia
Followers: 8

Kudos [?]: 276 [0], given: 20

### Show Tags

26 Sep 2009, 00:12
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!
Intern
Joined: 11 Aug 2009
Posts: 3
Followers: 0

Kudos [?]: 1 [0], given: 6

### Show Tags

29 Sep 2009, 02:26
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.
_________________

Manager
Joined: 14 Dec 2008
Posts: 169
Followers: 1

Kudos [?]: 24 [0], given: 39

### Show Tags

29 Sep 2009, 02:41
what is the solution to the original problem?..
Intern
Joined: 27 Aug 2009
Posts: 46
Followers: 4

Kudos [?]: 97 [1] , given: 1

### Show Tags

29 Sep 2009, 19:00
1
KUDOS
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
_________________

Salaries are low in recession. So, working for kudos now.

Intern
Joined: 07 Oct 2009
Posts: 19
Followers: 0

Kudos [?]: 10 [0], given: 0

### Show Tags

15 Oct 2009, 02:51
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.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13453
Followers: 575

Kudos [?]: 163 [0], given: 0

### Show Tags

15 Nov 2013, 05:04
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Permutation / combinations   [#permalink] 15 Nov 2013, 05:04
Similar topics Replies Last post
Similar
Topics:
The price ratio of oranges to lemons is 2:3. The price ratio of grapef 5 20 Oct 2016, 23:09
If the ratio of apples to bananas is 4 to 3 and the ratio of bananas 2 05 Apr 2016, 05:04
1 Six companies manufacture chocolates 3 19 Aug 2014, 20:18
13 The number of arrangement of letters of the word BANANA in 2 28 Oct 2013, 04:26
15 In the standard formulation of a flavored drink the ratio by 7 15 Nov 2010, 06:01
Display posts from previous: Sort by