Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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

1

This post received KUDOS

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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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

2

This post received KUDOS

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 [#permalink]
30 Oct 2012, 19:14

2

This post received KUDOS

Expert's post

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 _________________

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

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 [#permalink]
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!

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. _________________

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

Impossible is nothing to God.

gmatclubot

Re: In how many ways can 5 different candies be distributed in
[#permalink]
27 Dec 2012, 23:35