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In how many ways can 5 different candies be distributed in

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

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

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


:-D

Here, I don't see a shortcut, each case must be treated separately.
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Re: In how many ways can 5 different candies be distributed in [#permalink] New post 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.

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

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Re: In how many ways can 5 different candies be distributed in [#permalink] New post 27 Oct 2012, 23:07
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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.

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

Image Posted from GMAT ToolKit


Each one is explained in words (see text in blue above).
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Re: In how many ways can 5 different candies be distributed in [#permalink] New post 30 Oct 2012, 19:14
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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
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Last edited by GMATPill on 31 Oct 2012, 12:02, edited 1 time in total.
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Re: In how many ways can 5 different candies be distributed in [#permalink] New post 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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Re: In how many ways can 5 different candies be distributed in [#permalink] New post 31 Oct 2012, 12:02
Expert's post
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] New post 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] New post 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] 27 Dec 2012, 23:35
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