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Re: In how many ways can 5 different candies be distributed in
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Updated on: 31 Oct 2012, 13:02
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3
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!
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
Originally posted by GMATPill on 30 Oct 2012, 20:14.
Last edited by GMATPill on 31 Oct 2012, 13:02, edited 1 time in total.
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
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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.
Re: In how many ways can 5 different candies be distributed in
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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
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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...
Re: In how many ways can 5 different candies be distributed in
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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...
Re: In how many ways can 5 different candies be distributed in
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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!
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.)
Re: In how many ways can 5 different candies be distributed in
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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!
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.)
Re: In how many ways can 5 different candies be distributed in
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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. \(\frac{30}{2}=15\)
Re: In how many ways can 5 different candies be distributed in
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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!
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).
Re: In how many ways can 5 different candies be distributed in
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14 Apr 2015, 23:03
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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!
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.
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Re: In how many ways can 5 different candies be distributed in
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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!
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.
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Re: In how many ways can 5 different candies be distributed in
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09 Jul 2018, 11:37
BunuelVeritasPrepKarishma can you help me please? I am getting confused by these distribution problems. Can you help me differentiate the formulas I must use in each situation?
I had:
- for x different objects distributed among n people: n^x (which i thought was the case here in this question, and still didn't understand why not)
-for x different rings among n different fingers: use bars, number of bars = n-1, total ways = (x+bars)!/bars! (here, i didn't understand how fingers were treated differently from people, since in both cases they are different among themselves, that is, different fingers but also different people)
-for x identical objects among n different things: here as x also repeats, use formula as above but dividing by x!. that is, total ways= (x+bars)!/x! bars!
- for x identical objects among n different things, so that each n receives at least one object: x-1Cbars Here, for example, 20 fruits among 4 people, so that each receives at least one = 19C3= 19!/16!3! (here i can't see what the 16! would represent. in the case where people could get 0 fruits, you would divide by the repeated bars (3!) and the repeated fruits (20!), but i can't get what the 16! would be. is it because first you distribute 4 and then all other 16 would be repeated?
Thank you very much in advance. And thank you for all your genius posts, I am learning a lot from you both
gmatclubot
Re: In how many ways can 5 different candies be distributed in &nbs
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30% (01:17) correct 
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