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# In how many ways can 5 different marbles be distributed in 4

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Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
1
Kudos
I represent visually:

pockets = the space behind, after and between "/ and /"
marbles = 0

So we have to locate the marbles in each pocket for example:

00000/// => means all marbles in first pocket
/00000// => means all marbles in the second pocket
/000/0/0 => means no marbles in the first pocket, three marbles in the second pocket, one marble in the third pocket, and one marble int he fourth pocket
etc. I hope you get the idea.

So we have 00000 /// (8 objects total, 5 marbles plus 3 divisions for pockets) --> 8!/(5!*3!) = 56

Now, because we have identical pockets, we have 5 extra counts:

0000/// = 3 extra counts (because we have 3 empty spaces that can be occupied by the 5 marbles)
0000/0// = 2 extra counts (because we have 2 empty spaces that can be occupied by the 4 or 1 marble)

all other possible combinations do not allow repetitions, so:

56 - 3 - 2 = 51

If my logic is wrong, please show me. I am just trying to find a shortcut.

8!/(5!*3!) = 56
Here when you divide by 5!, you are saying that the marbles are identical. That is not correct here.
This method is used when identical objects are divided into distinct groups (discussed in the first link I gave above)

Further, logic of the following is not clear to me:
0000/// = 3 extra counts (because we have 3 empty spaces that can be occupied by the 5 marbles)
0000/0// = 2 extra counts (because we have 2 empty spaces that can be occupied by the 4 or 1 marble)
all other possible combinations do not allow repetitions
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Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
I represent visually:

pockets = the space behind, after and between "/ and /"
marbles = 0

So we have to locate the marbles in each pocket for example:

00000/// => means all marbles in first pocket
/00000// => means all marbles in the second pocket
/000/0/0 => means no marbles in the first pocket, three marbles in the second pocket, one marble in the third pocket, and one marble int he fourth pocket
etc. I hope you get the idea.

So we have 00000 /// (8 objects total, 5 marbles plus 3 divisions for pockets) --> 8!/(5!*3!) = 56

Now, because we have identical pockets, we have 5 extra counts:

0000/// = 3 extra counts (because we have 3 empty spaces that can be occupied by the 5 marbles)
0000/0// = 2 extra counts (because we have 2 empty spaces that can be occupied by the 4 or 1 marble)

all other possible combinations do not allow repetitions, so:

56 - 3 - 2 = 51

If my logic is wrong, please show me. I am just trying to find a shortcut.

8!/(5!*3!) = 56
Here when you divide by 5!, you are saying that the marbles are identical. That is not correct here.
This method is used when identical objects are divided into distinct groups (discussed in the first link I gave above)

Further, logic of the following is not clear to me:
0000/// = 3 extra counts (because we have 3 empty spaces that can be occupied by the 5 marbles)
0000/0// = 2 extra counts (because we have 2 empty spaces that can be occupied by the 4 or 1 marble)
all other possible combinations do not allow repetitions

I am trying to count how many marbles are over-counter in order to subtract it from the 56. Because there is no difference between the pockets, I suppose that the over-counter numbers are the ones when at least two empty pockets are consecutive.

When the five marbles are together there are 3 pockets without any ball. So there are 4 possible options with at least 2 empty spaces together. For the nature of the formula, I suppose that those are the over-counter from the 56. (It does not matter what pocket, so we can count one and leave out 3)

1) 00000 / / /
2) / 00000 / /
3) / / 0000 /
4) / / / 0000

That means that there are 3 unnecessary countings from the 56.

Likewise, in the count of 4 marbles in a pocket and 1 in another pocket, there are three scenarios where exist consecutive empty pockets

1) 0000 / 0 / /
2) / / 0000 / 0
3) 0000 / / 0

That means that there are 2 unnecessary countings from the 56

There is no other case where consecutive empty pockets exist.

The order does not matter, but the repetition of empty pockets does.

So, 56 - 2 - 3 = 51
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In how many ways can 5 different marbles be distributed in 4 [#permalink]
tusharGupta1 wrote:
In how many ways can 5 different marbles be distributed in 4 identical pockets?

a) 24

b) 51

c) 120

d) 625

e) 1024

Here are two posts you will find useful:

They discuss distributing
5 distinct things in 4 distinct groups
5 identical things in 4 distinct groups
5 distinct things in 4 identical groups (the case in point)
5 identical things in 4 identical groups

The first link you posted has extra characters in them. It is not working.

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In how many ways can 5 different marbles be distributed in 4 [#permalink]
utkmittal1 wrote:
tusharGupta1 wrote:
In how many ways can 5 different marbles be distributed in 4 identical pockets?

a) 24

b) 51

c) 120

d) 625

e) 1024

Hi

According to the problem we have to distribute 5 DIFFERENT marbles (say M1, M2, M3, M4, M5) into 4 IDENTICAL pockets. Since, the pockets are identical we need not worry about distributing the marbles to different pockets...we have to simply divide 5 marble into four or less groups. Therefore we can have following condition.

1. (5,0, 0, 0) --> 1 possibility (Becase the pockets are identical, there can be only one case. (5,0,0,0) and (0,5,0,0) are identical.

2. (4, 1, 0, 0) --->5C4 = 5 possibilities

3. (3, 2, 0, 0) ---> 5C3 = 10 possiblities

4. (3, 1, 1, 0) ----> 5C3 = 10 possiblities. There will be only 5C3 cases and not 5C3*2C1 because pockets are identical. For example. ((M1M2M3), M4, M5, 0) is same as ((M1M2M3), M4, 0, M5). Therefore just by grouping 3 out of 5 marbles we will get all the possibilities.

5. (2, 2, 1, 0) ---> 5C2*3C2/2 = 15 possibilities

6, (2, 1, 1, 1) --->5C2 = 10 possibilities

Total = 1+5+10+10+15+10 = 51 (Answer)

Note: we cannot form (1, 1, 1, 1, 1) because there are only 4 pockets.

why divided by 2 here??
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Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
KarishmaB Can questions like this big and calculation intensive come in the exam ?
PiyushK wrote:
I don't think such question will come in exam; it takes too much analysis case by case and definitely we have no short cut to solve this.

Posted from my mobile device
Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
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