|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 16 Feb 2011
Posts: 195
Schools: ABCD
|
In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
 21 Oct 2012, 18:00
1
This post received
KUDOS
8
This post was
BOOKMARKED
D
E
Question Stats:
28% (02:09) correct
72% (00:56) wrong based on 102 sessions
HideShow timer Statistics
In how many ways can 5 different candies be distributed in 4 identical baskets? (A) 120 (B) 51 (C) 24 (D) 5^4 (E) 4^5 Method1 (Long method): N= 0 fruits. 5-N-N-N : 5C5 = 1 4-1-N-N : 5C4 *1C1 = 5 3-2-N-N : 5C3*2C2 = 10 3-1-1-N : 5C3*2C1*1c1 = 20 2-2-1-N : 5C2*3C2*1C1 = 30 2-1-1-1 : 5C2*3*2*1 = 60 If I add these numbers, it doesn't equal to 51. What's my mistake? Thanks
Official Answer and Stats are available only to registered users. Register/ Login.
|
|
|
|
|
|
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
22 Oct 2012, 07:25
1
This post received
KUDOS
voodoochild wrote: In how many ways can 5 different candies be distributed in 4 identical baskets?
(A) 120
(B) 51
(C) 24
(D) 5^4
(E) 4^5
Method1 (Long method):
N= 0 fruits.
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : 5C3*2C1*1c1 = 20
2-2-1-N : 5C2*3C2*1C1 = 30
2-1-1-1 : 5C2*3*2*1 = 60
If I add these numbers, it doesn't equal to 51. What's my mistake?
Thanks 5-N-N-N : 5C5 = 1 - OK 4-1-N-N : 5C4 *1C1 = 5 - OK 3-2-N-N : 5C3*2C2 = 10 - OK 3-1-1-N : 5C3*2C1*1C1 = 20 - NO - only 10, once you decide to split the remaining two after choosing the 3 before, no factor of 2 needed 2-2-1-N : 5C2*3C2*1C1 = 30 - NO - it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first 2-1-1-1 : 5C2*3*2*1 = 60 - NO - only 10, because you only choose those 2 to be placed together, all the other three you put in different baskets, no choices Total of 1 + 5 + 10 + 10 + 15 + 10 = 51. Answer B.
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
|
|
|
|
|
|
Manager
Joined: 16 Feb 2011
Posts: 195
Schools: ABCD
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
22 Oct 2012, 09:03
Thanks Eva. A quick question - is there any shortcut method for this question? Appreciate your help. If GMATClub had allowed, I would have given you 100 kudos. Thanks for helping me.
Thanks
|
|
|
|
|
|
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
22 Oct 2012, 09:17
1
This post received
KUDOS
voodoochild wrote: Thanks Eva. A quick question - is there any shortcut method for this question? Appreciate your help. If GMATClub had allowed, I would have given you 100 kudos. Thanks for helping me.
Thanks  Here, I don't see a shortcut, each case must be treated separately.
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
|
|
|
|
|
|
Intern
Joined: 22 Oct 2012
Posts: 19
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
27 Oct 2012, 21:44
EvaJager wrote:
5-N-N-N : 5C5 = 1 - OK
4-1-N-N : 5C4 *1C1 = 5 - OK
3-2-N-N : 5C3*2C2 = 10 - OK
3-1-1-N : 5C3*2C1*1C1 = 20 - NO - only 10, once you decide to split the remaining two after choosing the 3 before, no factor of 2 needed
2-2-1-N : 5C2*3C2*1C1 = 30 - NO - it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first
2-1-1-1 : 5C2*3*2*1 = 60 - NO - only 10, because you only choose those 2 to be placed together, all the other three you put in different baskets, no choices
Total of 1 + 5 + 10 + 10 + 15 + 10 = 51.
Answer B.
Hi, Sorry, i have a hard time with this topic, Do you mean that for 3-1-1-N we don t need 2C1? 2-2-1-N we don t need 3C1? I stll don t understand well the reason... Could you please explain it in other words? Thank you R26  Posted from GMAT ToolKit
|
|
|
|
|
|
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
28 Oct 2012, 00:07
2
This post received
KUDOS
R26 wrote: EvaJager wrote:
5-N-N-N : 5C5 = 1 - OK
4-1-N-N : 5C4 *1C1 = 5 - OK
3-2-N-N : 5C3*2C2 = 10 - OK
3-1-1-N : 5C3*2C1*1C1 = 20 - NO - only 10, once you decide to split the remaining two after choosing the 3 before, no factor of 2 needed
2-2-1-N : 5C2*3C2*1C1 = 30 - NO - it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first
2-1-1-1 : 5C2*3*2*1 = 60 - NO - only 10, because you only choose those 2 to be placed together, all the other three you put in different baskets, no choices
Total of 1 + 5 + 10 + 10 + 15 + 10 = 51.
Answer B.
Hi, Sorry, i have a hard time with this topic, Do you mean that for 3-1-1-N we don t need 2C1? 2-2-1-N we don t need 3C1? I stll don t understand well the reason... Could you please explain it in other words? Thank you R26 Posted from GMAT ToolKitEach one is explained in words (see text in blue above).
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
|
|
|
|
|
|
SVP
Joined: 14 Apr 2009
Posts: 2085
Location: New York, NY
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
30 Oct 2012, 20:14
2
This post received
KUDOS
Evajager's correct.
As further clarification - whenever you have 2 baskets with exactly the same number of candies, swapping those two baskets does NOT count has two counts of distribution. Yet, when you multiply the 5C3 * 2C1 * 1C1 -> that counts the swapping of the two baskets as 2 separate counts.
Therefore, in order to account for the fact that order does not matter (that these situations should not be double counted) - you should divide by the # of baskets that are duplicate. So in this case, you would need to divide by 2!
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10
2-2-1-N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15
2-1-1-1 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2-1-1-1: you should divide by 3! = 6 because you have 3 identical baskets with 1 candy each that are counted as duplicates but really they should all count as 1.
So in total, we have 1 + 5 + 10 + 10 + 15 + 10 = 51
Last edited by GMATPill on 31 Oct 2012, 13:02, edited 1 time in total.
|
|
|
|
|
|
Intern
Joined: 22 Oct 2012
Posts: 19
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
30 Oct 2012, 23:04
gmatpill wrote: Evajager's correct.
As further clarification - whenever you have 2 baskets with exactly the same number of candies, swapping those two baskets does NOT count has two counts of distribution. Yet, when you multiply the 5C3 * 2C1 * 1C1 -> that counts the swapping of the two baskets as 2 separate counts.
Therefore, in order to account for the fact that order does not matter (that these situations should not be double counted) - you should divide by the # of baskets that are duplicate. So in this case, you would need to divide by 2!
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10
2-2-1-N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15
2-1-1-1 : (5C2*3*2*1) / (6!) = 60 / 6 = 10
In the last set up: 2-1-1-1: you should divide by 3! = 6 because you have 3 identical baskets with 1 candy each that are counted as duplicates but really they should all count as 1.
So in total, we have 1 + 5 + 10 + 10 + 15 + 10 = 51 Thank you very much, it is clear now ^_^ (By the way, is it a typo in red for the 2-1-1-1? You wanted to say 3! But though directly of the result being 6.) Thank you again all of you R26
|
|
|
|
|
|
SVP
Joined: 14 Apr 2009
Posts: 2085
Location: New York, NY
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
31 Oct 2012, 13:02
R26 wrote: gmatpill wrote: Evajager's correct.
As further clarification - whenever you have 2 baskets with exactly the same number of candies, swapping those two baskets does NOT count has two counts of distribution. Yet, when you multiply the 5C3 * 2C1 * 1C1 -> that counts the swapping of the two baskets as 2 separate counts.
Therefore, in order to account for the fact that order does not matter (that these situations should not be double counted) - you should divide by the # of baskets that are duplicate. So in this case, you would need to divide by 2!
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10
2-2-1-N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15
2-1-1-1 : (5C2*3*2*1) / (6!) = 60 / 6 = 10
In the last set up: 2-1-1-1: you should divide by 3! = 6 because you have 3 identical baskets with 1 candy each that are counted as duplicates but really they should all count as 1.
So in total, we have 1 + 5 + 10 + 10 + 15 + 10 = 51 Thank you very much, it is clear now ^_^ (By the way, is it a typo in red for the 2-1-1-1? You wanted to say 3! But though directly of the result being 6.) Thank you again all of you R26 Oh yes, I edited to correct the 6! to 3! = 6.
|
|
|
|
|
|
Director
Joined: 29 Nov 2012
Posts: 878
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
24 Dec 2012, 08:04
Is this the only way to solve this problem? tricky question I misunderstood it completely..
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
|
|
|
|
|
|
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GPA: 3.23
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
28 Dec 2012, 00:35
voodoochild wrote: In how many ways can 5 different candies be distributed in 4 identical baskets?
(A) 120
(B) 51
(C) 24
(D) 5^4
(E) 4^5
We just determine the distribution of specific candies to a grouping of 5,4,2,3, or 1 per basket but there is no need to arrange the distribution on to the baskets since the containers are identical.5-0-0-0: \(\frac{5!}{5!0!} = 1\) 4-1-0-0: \(\frac{5!}{4!1!}*1 = 5\) 3-2-0-0: \(\frac{5!}{3!2!}*\frac{2!}{2!} = 10\) 3-1-1-0: \(\frac{5!}{3!2!}*\frac{2!}{1!} * \frac{1!}{1!} = 20\) Then, we divide by 2! since 1 and 1 even interchanged doesn't matter. \(\frac{20}{2}=10\) 2-1-1-1: \(\frac{5!}{2!3!}*\frac{3!}{1!2!}*\frac{2!}{1!1!}*1=60\) Then, we divide by 3! since 1,1 and 1 even interchanged doesn't matter since the baskets are identical. \(\frac{60}{3!}=10\) 2-2-1-0: \(\frac{5!}{2!3!}*\frac{3!}{2!1!}=30\) Then divide by 2! since 2 and 2 even interchanged doesn't matter. \(\frac{30}{2}=15\) Answer: \(1+5+10+10+10+15=51\)
_________________
Impossible is nothing to God.
|
|
|
|
|
|
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16024
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
13 Apr 2015, 10:38
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.
_________________
GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources
|
|
|
|
|
|
Intern
Joined: 06 Apr 2015
Posts: 7
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
14 Apr 2015, 20:24
GMATPill wrote: Evajager's correct.
As further clarification - whenever you have 2 baskets with exactly the same number of candies, swapping those two baskets does NOT count has two counts of distribution. Yet, when you multiply the 5C3 * 2C1 * 1C1 -> that counts the swapping of the two baskets as 2 separate counts.
Therefore, in order to account for the fact that order does not matter (that these situations should not be double counted) - you should divide by the # of baskets that are duplicate. So in this case, you would need to divide by 2!
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10
2-2-1-N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15
2-1-1-1 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2-1-1-1: you should divide by 3! = 6 because you have 3 identical baskets with 1 candy each that are counted as duplicates but really they should all count as 1.
So in total, we have 1 + 5 + 10 + 10 + 15 + 10 = 51 I dont quite understand why we are multiplying combinatorics (ie 5C3*2C2, etc.)? Is there some general rule or application on when I should do this? Also in the last sequence 2-1-1-1, you do 5C2 and then 3! which I'm assuming is just 3C1. Why isn't it 5C2*3C1*2C1*1C1? (Appreciate that answering the above may explain this). Thanks in advance
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
14 Apr 2015, 23:03
Sneakysam wrote: GMATPill wrote: Evajager's correct.
As further clarification - whenever you have 2 baskets with exactly the same number of candies, swapping those two baskets does NOT count has two counts of distribution. Yet, when you multiply the 5C3 * 2C1 * 1C1 -> that counts the swapping of the two baskets as 2 separate counts.
Therefore, in order to account for the fact that order does not matter (that these situations should not be double counted) - you should divide by the # of baskets that are duplicate. So in this case, you would need to divide by 2!
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10
2-2-1-N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15
2-1-1-1 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2-1-1-1: you should divide by 3! = 6 because you have 3 identical baskets with 1 candy each that are counted as duplicates but really they should all count as 1.
So in total, we have 1 + 5 + 10 + 10 + 15 + 10 = 51 I dont quite understand why we are multiplying combinatorics (ie 5C3*2C2, etc.)? Is there some general rule or application on when I should do this? Also in the last sequence 2-1-1-1, you do 5C2 and then 3! which I'm assuming is just 3C1. Why isn't it 5C2*3C1*2C1*1C1? (Appreciate that answering the above may explain this). Thanks in advance When two selection are made for the same case, we multiply (the AND rule) Say, in how many ways can you select a meal which includes a burger and a beverage. There are 2 different kinds of burgers and 3 different kinds of beverages. You can do it in 2C1 * 3C1 ways. You have to select a burger AND a beverage so you multiply their respective selections and get the answer as 6. On the other hand, in how many ways can you select a meal of either a burger or a pizza but not both? There are 2 different kinds of burgers and 3 different kinds of pizzas. You can do this in 2C1 + 3C1 = 5 ways. You add here because you have to select burger OR pizza. When you have OR, you add. Here you have to select 3 candies for one basket AND 2 candies for another basket so you do 5C3 * 2C2 Yes, 3*2*1 at the end is 3C1*2C1*1C1.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews
|
|
|
|
|
|
Director
Joined: 07 Aug 2011
Posts: 579
Concentration: International Business, Technology
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
15 Apr 2015, 03:54
VeritasPrepKarishma wrote: Sneakysam wrote: GMATPill wrote: Evajager's correct.
As further clarification - whenever you have 2 baskets with exactly the same number of candies, swapping those two baskets does NOT count has two counts of distribution. Yet, when you multiply the 5C3 * 2C1 * 1C1 -> that counts the swapping of the two baskets as 2 separate counts.
Therefore, in order to account for the fact that order does not matter (that these situations should not be double counted) - you should divide by the # of baskets that are duplicate. So in this case, you would need to divide by 2!
5-N-N-N : 5C5 = 1
4-1-N-N : 5C4 *1C1 = 5
3-2-N-N : 5C3*2C2 = 10
3-1-1-N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10
2-2-1-N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15
2-1-1-1 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2-1-1-1: you should divide by 3! = 6 because you have 3 identical baskets with 1 candy each that are counted as duplicates but really they should all count as 1.
So in total, we have 1 + 5 + 10 + 10 + 15 + 10 = 51 I dont quite understand why we are multiplying combinatorics (ie 5C3*2C2, etc.)? Is there some general rule or application on when I should do this? Also in the last sequence 2-1-1-1, you do 5C2 and then 3! which I'm assuming is just 3C1. Why isn't it 5C2*3C1*2C1*1C1? (Appreciate that answering the above may explain this). Thanks in advance When two selection are made for the same case, we multiply (the AND rule) Say, in how many ways can you select a meal which includes a burger and a beverage. There are 2 different kinds of burgers and 3 different kinds of beverages. You can do it in 2C1 * 3C1 ways. You have to select a burger AND a beverage so you multiply their respective selections and get the answer as 6. On the other hand, in how many ways can you select a meal of either a burger or a pizza but not both? There are 2 different kinds of burgers and 3 different kinds of pizzas. You can do this in 2C1 + 3C1 = 5 ways. You add here because you have to select burger OR pizza. When you have OR, you add. Here you have to select 3 candies for one basket AND 2 candies for another basket so you do 5C3 * 2C2 Yes, 3*2*1 at the end is 3C1*2C1*1C1. i am not sure why the below approach is not working for this one .. let say there are 4 Kids , and we want to distribute 5candies among them . CCCCC||| is one possible arrangement , here the first kid got all the 5 candies. so why cant we use \(^8C_3\) to get 56 as answer.
_________________
Thanks,
Lucky
_______________________________________________________
Kindly press the  to appreciate my post !! 
|
|
|
|
|
|
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16024
|
Re: In how many ways can 5 different candies be distributed in [#permalink]
Show Tags
20 Apr 2016, 12:00
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.
_________________
GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources
|
|
|
|
|
|
|
Re: In how many ways can 5 different candies be distributed in
[#permalink]
20 Apr 2016, 12:00
|
|
|
|
|
|
|
|
|
|
|
|
|
2
