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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

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