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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. 5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : 5C3*2C1*1c1 = 20 221N : 5C2*3C2*1C1 = 30 2111 : 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]
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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.
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : 5C3*2C1*1c1 = 20 221N : 5C2*3C2*1C1 = 30 2111 : 5C2*3*2*1 = 60
If I add these numbers, it doesn't equal to 51. What's my mistake?
Thanks 5NNN : 5C5 = 1  OK 41NN : 5C4 *1C1 = 5  OK 32NN : 5C3*2C2 = 10  OK 311N : 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 221N : 5C2*3C2*1C1 = 30  NO  it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first 2111 : 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]
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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.
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Re: In how many ways can 5 different candies be distributed in [#permalink]
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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]
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27 Oct 2012, 20:44
EvaJager wrote: 5NNN : 5C5 = 1  OK 41NN : 5C4 *1C1 = 5  OK 32NN : 5C3*2C2 = 10  OK 311N : 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 221N : 5C2*3C2*1C1 = 30  NO  it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first 2111 : 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 311N we don t need 2C1? 221N 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]
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R26 wrote: EvaJager wrote: 5NNN : 5C5 = 1  OK 41NN : 5C4 *1C1 = 5  OK 32NN : 5C3*2C2 = 10  OK 311N : 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 221N : 5C2*3C2*1C1 = 30  NO  it is 15 = 30/2; baskets being identical, doesn't matter which group of 2 you choose first 2111 : 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 311N we don t need 2C1? 221N 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]
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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!
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10 221N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15 2111 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2111: 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.



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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!
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10 221N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15 2111 : (5C2*3*2*1) / (6!) = 60 / 6 = 10
In the last set up: 2111: 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 2111? 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]
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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!
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10 221N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15 2111 : (5C2*3*2*1) / (6!) = 60 / 6 = 10
In the last set up: 2111: 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 2111? 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.5000: \(\frac{5!}{5!0!} = 1\) 4100: \(\frac{5!}{4!1!}*1 = 5\) 3200: \(\frac{5!}{3!2!}*\frac{2!}{2!} = 10\) 3110: \(\frac{5!}{3!2!}*\frac{2!}{1!} * \frac{1!}{1!} = 20\) Then, we divide by 2! since 1 and 1 even interchanged doesn't matter. \(\frac{20}{2}=10\) 2111: \(\frac{5!}{2!3!}*\frac{3!}{1!2!}*\frac{2!}{1!1!}*1=60\) Then, we divide by 3! since 1,1 and 1 even interchanged doesn't matter since the baskets are identical. \(\frac{60}{3!}=10\) 2210: \(\frac{5!}{2!3!}*\frac{3!}{2!1!}=30\) 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!
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10 221N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15 2111 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2111: 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 2111, 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



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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
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!
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10 221N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15 2111 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2111: 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 2111, 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.
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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!
5NNN : 5C5 = 1 41NN : 5C4 *1C1 = 5 32NN : 5C3*2C2 = 10 311N : (5C3*2C1*1c1) (2!) = 20 / 2 = 10 221N : (5C2*3C2*1C1) / (2!) = 30 / 2 = 15 2111 : (5C2*3*2*1) / (3!) = 60 / 6 = 10
In the last set up: 2111: 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 2111, 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.
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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/howmanyway ... 37198.htmlEach marble has 4 options so 4^5?




Re: In how many ways can 5 different candies be distributed in
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