How many ways can you distribute 5 marbles in 3 baskets

[hairen]

How many ways can you distribute 5 marbles in 3 identical baskets such that each baskets has at least 1 marble. I'm trying to solve this question using the "slot method" (the slot method is the method MGMAT advises to use for solving combinatorics questions, you can google it). I'm having some trouble arriving at the right answer. Please advise me on what I'm doing wrong.
Here is my approach:
1. Create a slot for each decision
_ _ / _ _ / _ OR _ _ _ / _ / _
2/2/1 or 3/1/1
ie: we can have 2 marbles in 2 baskets and 1 marble in 1 basket OR 3 marbles in a basket and 1 marble in 2 baskets.
2. Feel in the slots with the number of options
5 4 3 2 1 OR 5 4 3 2 1
3. Multiply
4. Divide by the factorial(s) of the number of interchangeable set(s)

I'm not sure how to proceed after step 1

[hkparikh09]

I got 6 as the answer, is that the correct answer? I'm not too confident, but here is what I did:

I could think of 2 main arrangements to satisfy both requirements, the same that you came up with:

3,1,1 and 2,2,1

Next, I said that the two arrangements above could be manipulated by "switching the slots around." For example, the first arrangement could be switched from 3,1,1 to 1,1,3 or 1,3,1. You can do the same for the second arrangement.

2,2,1 or 1,2,2 or 2,1,2

So a total of 6 arrangements

Another way is:

For the 1st arrangement, you have three baskets, two with the same number of marbles: 3! / 2! = 3.

For the 2nd arrangement, you have three baskets, two with the same number of marbles: 3! / 2! = 3.

Arrangement one or arrangement two: 3 + 3 = 6.

[hairen]

Hi hkparikh09,

The answer is 25. Here is a solution using the formulas:

For 3_1_1
(5C3*2C1*1C1)/2! = 10

For 2_2_1
(5C2*3C2*1C1)/2! = 15

10+15=25

If somebody could explain how to solve this problem using the slot method, that would be great.

Ron?

[RonPurewal]

so yeah, if you put this problem in front of me, i'd just make a list.

* if the marbles are distributed 2/2/1:
12/34/5
12/35/4
12/45/3
13/24/5
13/25/4
13/45/2
14/23/5
14/25/3
14/35/2
15/23/4
15/24/3
15/34/2
23/45/1
24/35/1
25/34/1
... 15 ways to do it that way.

* if they are distributed 3/1/1:
123/4/5
124/3/5
125/3/4
134/2/5
135/2/4
145/2/3
234/1/5
235/1/4
245/1/3
345/1/2
... 10 ways to do it this way.

total 25. that was not very painful, nor was it slow.

--

slot method:
if it's "_ _ / _ _ / _"
then ...
the first 2 slots are interchangeable ("order doesn't matter"), so that's (5 x 4)/2! = 10.
the next 2 slots are also interchangeable, so that's (3 x 2)/2! = 3.
but notice that the first pair of slots is also interchangeable with the second pair of slots, since the baskets are identical; e.g., 12/34/5 is the same arrangement as 34/12/5. so you have to divide by 2! again to account for this: (3 x 10)/2! = 15.

if it's "_ _ _ / _ / _"
then all you have to worry about is the first three slots, because, once you've picked those, the contents of the other two (singleton) slots are automatic.
this is your standard choice of three things out of 5 where order doesn't matter: (5 x 4 x 3)/3! = 10.

so 10 + 15 = 25.

but, that's icky and gross, and it smells bad, too. i'd rather just make a list.

[hairen]

Thanks a lot Ron! this is super helpful.

I have a clarifying question for the 3_1_1 scenario tho.
Why are we not multiplying the last two slots by 2 and 1.
ie: why is it not (5 x 4 x 3 x 2 x 1 )/3!

thanks in advance

[RonPurewal]

Thanks a lot Ron! this is super helpful.

I have a clarifying question for the 3_1_1 scenario tho.
Why are we not multiplying the last two slots by 2 and 1.
ie: why is it not (5 x 4 x 3 x 2 x 1 )/3!

thanks in advance

in that scenario, the two single slots are equivalent -- remember, the buckets are indistinguishable -- so there are no further distinctions.

you could multiply by 2 and 1 ... but then you would have to divide by 2! afterward, for the same reason as the last step in the "_ _ / _ _ / _" situation.

... or you could just make a list. and then not have to worry about it!

[tmipanthers]

so yeah, if you put this problem in front of me, i'd just make a list.

* if the marbles are distributed 2/2/1:
12/34/5
12/35/4
12/45/3
13/24/5
13/25/4
13/45/2
14/23/5
14/25/3
14/35/2
15/23/4
15/24/3
15/34/2
23/45/1
24/35/1
25/34/1
... 15 ways to do it that way.

* if they are distributed 3/1/1:
123/4/5
124/3/5
125/3/4
134/2/5
135/2/4
145/2/3
234/1/5
235/1/4
245/1/3
345/1/2
... 10 ways to do it this way.

total 25. that was not very painful, nor was it slow.

--

slot method:
if it's "_ _ / _ _ / _"
then ...
the first 2 slots are interchangeable ("order doesn't matter"), so that's (5 x 4)/2! = 10.
the next 2 slots are also interchangeable, so that's (3 x 2)/2! = 3.
but notice that the first pair of slots is also interchangeable with the second pair of slots, since the baskets are identical; e.g., 12/34/5 is the same arrangement as 34/12/5. so you have to divide by 2! again to account for this: (3 x 10)/2! = 15.

if it's "_ _ _ / _ / _"
then all you have to worry about is the first three slots, because, once you've picked those, the contents of the other two (singleton) slots are automatic.
this is your standard choice of three things out of 5 where order doesn't matter: (5 x 4 x 3)/3! = 10.

so 10 + 15 = 25.

but, that's icky and gross, and it smells bad, too. i'd rather just make a list.

Hi Ron. Thank you for the explanation. However, I am still very confused.

List: How did you make this list? I have no idea what the numbers represent i.e. (12/34/5)

Slot Method:

I understand that for 3-1-1 it's (5C3*2C1*1C1) and for the 2-2-1 (5C2*3C2*1C1), and then add them both up. But I don't understand why they both are first divided by 2?

Further more, I see that I lack some basic skills to understand this. I have poured over the 5 MGMAT math books except for the advanced ones. I feel i should be able to understand these problems already. If not initially, at least after further explanation. Yet I am still not there. What should I do?

[hariready]

so yeah, if you put this problem in front of me, i'd just make a list.

* if the marbles are distributed 2/2/1:

... 15 ways to do it that way.

* if they are distributed 3/1/1:
... 10 ways to do it this way.

total 25. that was not very painful, nor was it slow..

Hi Ron!
There are 5 marbles and 3 Identical baskets right?
So it shouldn't matter which basket you pick, because the baskets are identical. Makes sense?
I think only the different combinations of the marbles should be taken into account not the selection of baskets. So it will be 2X(3!/2!) = 6?

[RonPurewal]

There are 5 marbles and 3 Identical baskets right?
So it shouldn't matter which basket you pick, because the baskets are identical. Makes sense?

my approach above is based on this idea.

i'm not seeing where your math comes from; perhaps you could explain. but, i made an exhaustive list of all the possibilities above, so that's solid proof that there are 25 ways... you can count them!