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In how many ways can 5 different candies be distributed in [#permalink]
 21 Oct 2012, 18:00
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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
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Re: In how many ways can 5 different candies be distributed in [#permalink]
22 Oct 2012, 07:25
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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]
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
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Re: In how many ways can 5 different candies be distributed in [#permalink]
22 Oct 2012, 09:17
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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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Re: In how many ways can 5 different candies be distributed in [#permalink]
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
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Re: In how many ways can 5 different candies be distributed in [#permalink]
28 Oct 2012, 00:07
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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 Posted from GMAT ToolKitEach 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]
30 Oct 2012, 20:14
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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, 13:02, edited 1 time in total.
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Re: In how many ways can 5 different candies be distributed in [#permalink]
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
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Re: In how many ways can 5 different candies be distributed in [#permalink]
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.
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Re: In how many ways can 5 different candies be distributed in [#permalink]
24 Dec 2012, 08:04
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]
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!} = 14-1-0-0: \frac{5!}{4!1!}*1 = 53-2-0-0: \frac{5!}{3!2!}*\frac{2!}{2!} = 103-1-1-0: \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}=102-1-1-1: \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!}=102-2-1-0: \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}=15Answer: 1+5+10+10+10+15=51
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Re: In how many ways can 5 different candies be distributed in
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28 Dec 2012, 00:35
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