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# In how many ways can 5 different candies be distributed in

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Manager
Joined: 16 Feb 2011
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Schools: ABCD
In how many ways can 5 different candies be distributed in [#permalink]

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21 Oct 2012, 17:00
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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
[Reveal] Spoiler: OA
Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)
Re: In how many ways can 5 different candies be distributed in [#permalink]

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22 Oct 2012, 06:25
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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.

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Manager
Joined: 16 Feb 2011
Posts: 195
Schools: ABCD
Re: In how many ways can 5 different candies be distributed in [#permalink]

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22 Oct 2012, 08:03
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: 608
WE: Science (Education)
Re: In how many ways can 5 different candies be distributed in [#permalink]

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22 Oct 2012, 08:17
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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.
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Intern
Joined: 22 Oct 2012
Posts: 15
Re: In how many ways can 5 different candies be distributed in [#permalink]

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27 Oct 2012, 20:44
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.

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: 608
WE: Science (Education)
Re: In how many ways can 5 different candies be distributed in [#permalink]

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27 Oct 2012, 23:07
2
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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.

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

Each one is explained in words (see text in blue above).
_________________

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Love GMAT Quant questions and running.

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Joined: 14 Apr 2009
Posts: 2146
Location: New York, NY
Re: In how many ways can 5 different candies be distributed in [#permalink]

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30 Oct 2012, 19:14
2
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This post was
BOOKMARKED
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, 12:02, edited 1 time in total.
Intern
Joined: 22 Oct 2012
Posts: 15
Re: In how many ways can 5 different candies be distributed in [#permalink]

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30 Oct 2012, 22:04
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
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Joined: 14 Apr 2009
Posts: 2146
Location: New York, NY
Re: In how many ways can 5 different candies be distributed in [#permalink]

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31 Oct 2012, 12:02
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.
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Re: In how many ways can 5 different candies be distributed in [#permalink]

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24 Dec 2012, 07:04
Is this the only way to solve this problem? tricky question I misunderstood it completely..
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Re: In how many ways can 5 different candies be distributed in [#permalink]

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27 Dec 2012, 23:35
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: $$1+5+10+10+10+15=51$$
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Re: In how many ways can 5 different candies be distributed in [#permalink]

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14 Apr 2015, 19:24
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).

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Re: In how many ways can 5 different candies be distributed in [#permalink]

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14 Apr 2015, 22:03
Expert's post
1
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BOOKMARKED
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).

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.
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Re: In how many ways can 5 different candies be distributed in [#permalink]

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15 Apr 2015, 02:54
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).

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.
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Re: In how many ways can 5 different candies be distributed in [#permalink]

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21 Sep 2017, 12:25
Why not E like here? https://gmatclub.com/forum/how-many-way ... 37198.html
Each marble has 4 options so 4^5?
Re: In how many ways can 5 different candies be distributed in   [#permalink] 21 Sep 2017, 12:25
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# In how many ways can 5 different candies be distributed in

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