In how many ways can 5 different candies be distributed in

Question

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

GMATPill · Accepted Answer

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

EvaJager · Answer

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.

voodoochild · Answer

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

EvaJager · Answer

Here, I don't see a shortcut, each case must be treated separately.

R26 · Answer

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

https://gmatclub.com/static/posted_from.png     Posted from    GMAT ToolKit

EvaJager · Answer

Each one is explained in words (see text in blue above).

R26 · Answer

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

GMATPill · Answer

Oh yes, I edited to correct the 6! to 3! = 6.

fozzzy · Answer

Is this the only way to solve this problem? tricky question I misunderstood it completely..

mbaiseasy · Answer

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

Sneakysam · Answer

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

KarishmaB · Answer

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.

Lucky2783 · Answer

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.

inter · Answer

Why not E like here? https://gmatclub.com/forum/how-many-way ... 37198.html Each marble has 4 options so 4^5?

Juliaz · Answer

Bunuel VeritasPrepKarishma 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

GMATPanther · Answer

why cant it be 4^5 ? whats different https://gmatclub.com/forum/in-how-many- ... 31187.html ? tx

Fdambro294 · Answer

Since the Candies are Different ---> but the Baskets they are Distributed to are IDENTICAL, it only matters which candies are chosen to end up in a particular Number Stack. It does not matter which Basket gets that Number Stack.

There are 6 Different Scenarios in which the 5 Different Candies can be Divided among the 4 Identical Baskets:

Scenario 1: 0 - 0 - 0 - 5

"5 Choose 5" = 1 Way in which we can choose the 5 Different Candies. It does NOT matter which Basket receives them.

Scenario 2: 0 - 0 - 1 - 4
"5 Choose 4" and "1 Choose 1" = 5! / 4! * 1! = 5 Ways in which we can choose the 4 Different Candies for 1 Stack and 1 Candy for the Other Stack.

Scenario 3: 0 - 0 - 2 - 3
"5 Choose 3" and "2 Choose 2" = 5! / 3! * 2! = 10 Ways in which we can choose the 3 Different Candies for 1 Stack and 2 Candies for the Other Stack

Scenario 4: 0 - 1 - 1 - 3
"5 choose 3" and "2 Choose 1" and "1 choose 1" =
5! / 3! * 1! * 1! = 10 Ways

Scenario 5: 0 - 1 - 2 - 2
"5 Choose 2" and "3 Choose 2" and "1 Choose 1" =

However, since 2 of the Identical Stacks are of EQUAL SIZE (2 and 2), we will have Over-counting. For Each 1 Distribution we want counted, 2! will actually be Counted. Therefore, we must Divide by 2! to Remove these Repeats.

5! / 2! * 2! * 1! * (1 / 2!) = 5! / 2 * 2 * 2 = 5 * 4 * 3 / 2 * 2 = 15 Ways

Scenario 6: 1 - 1 - 1 - 2
"5 Choose 2" = 5! / 2! * 3! = 10 Ways

Adding up the 6 Scenarios =

1 + 5 + 10 + 10 + 15 + 10 = 51 Different Ways in which we can Distribute 5 Different Candies to 4 Identical Baskets

kkannan2 · Answer

Hi EvaJager , I saw that you had given an explanation for this similar distribution question as well: https://gmatclub.com/forum/how-many-way ... 37198.html What is the difference between the wording of this question and the other linked question? I don't see anything significant and want to improve my understanding of such questions. Could you please elaborate why this solution is much more involved?

kkannan2 · Answer

Hi  Bunuel  and  KarishmaB , I'd like to +1 the post from  Juliaz . It is indeed confusing when to apply the methods used for these kinds of distribution problems. I have provided an identical example above that has a much simpler solution path, so I'm also trying to better understand when to use what approach to solve these problems efficiently.

To this end, I think a breakdown of when to use the calculation tactics mentioned in in Julia's post can help us all understand this a bit better. I am also curious if Julia's list is exhaustive or if there are other calculation techniques that we should be aware of. Please let us know if you can help!

Thanks, Kaushik

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