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Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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28 Apr 2014, 10:58

3

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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.

Last edited by utkmittal1 on 14 Jun 2014, 05:49, edited 1 time in total.

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
_________________

In how many ways can 5 different marbles be distributed in 4 identical pockets?

a) 24

b) 51

c) 120

d) 625

e) 1024

The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets.

5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how.

{5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits.

{4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits.

{3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits.

{3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits.

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

{2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits.

Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways
_________________

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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30 May 2014, 14:46

VeritasPrepKarishma 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

The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets.

5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how.

{5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits.

{4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits.

{3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits.

{3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits.

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

{2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits.

Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways

Hi Karishma,

I read your posts but still it isn't clear to me how one needs to count each event instead of just applying a combinatorics formula such as 'n+r-1 C r-1'?

Bunuel, you are also invited to this discussion since I basically stole this formula from you

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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01 Jun 2014, 11:18

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.
_________________

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In how many ways can 5 different marbles be distributed in 4 identical pockets?

a) 24

b) 51

c) 120

d) 625

e) 1024

The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets.

5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how.

{5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits.

{4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits.

{3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits.

{3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits.

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

{2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits.

Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways

Hi Karishma,

I read your posts but still it isn't clear to me how one needs to count each event instead of just applying a combinatorics formula such as 'n+r-1 C r-1'?

Bunuel, you are also invited to this discussion since I basically stole this formula from you

Many thanks Cheers J

(n+r-1)C(r-1) gives the number of ways of distributing n identical things in r distinct groups - this questions distributes n distinct marbles in r identical groups.

For identical groups, you need to count.
_________________

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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14 Jun 2014, 03:53

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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 = 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.

I like your detailed explanation. However, calculation for case no 5 seems incorrect. It would give 30. We need to divide by 2 to avoid duplicating cases like M1 M2 , M3 M4 , M5 , 0 and M3 M4 , M1 M2, M5 , 0 since the groups are identical.

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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14 Jun 2014, 05:48

ValKilmer wrote:

I like your detailed explanation. However, calculation for case no 5 seems incorrect. It would give 30. We need to divide by 2 to avoid duplicating cases like M1 M2 , M3 M4 , M5 , 0 and M3 M4 , M1 M2, M5 , 0 since the groups are identical.

Thanks Arpit

Hi ValKilmiler

Thanks for correcting me. Don't know how I missed to divide it by 2. Updated my post.

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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06 Nov 2014, 02:33

VeritasPrepKarishma 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

The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets.

5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how.

{5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits.

{4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits.

{3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits.

{3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits.

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

{2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits.

Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways

Hi Karishma,

Could you please provide your comments on following scenario. I am getting 30 in this instead of 15

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

Could you please provide your comments on following scenario. I am getting 30 in this instead of 15

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

Thanks

As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to un-arrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15

If you present your calculation, I can tell you what you are missing.
_________________

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]

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04 Sep 2015, 15:03

VeritasPrepKarishma wrote:

PathFinder007 wrote:

Hi Karishma,

Could you please provide your comments on following scenario. I am getting 30 in this instead of 15

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

Thanks

As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to un-arrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15

If you present your calculation, I can tell you what you are missing.

I am getting 30 as well, here is my reasoning: [2 2 1 0] - I can select the first 2 out of 5 in 5C2=10 ways - for the second 2 I can select 2 out of 3 3C2= 3 way Multiplying 10*3=30

Could you please provide your comments on following scenario. I am getting 30 in this instead of 15

{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.

Thanks

As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to un-arrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15

If you present your calculation, I can tell you what you are missing.

I am getting 30 as well, here is my reasoning: [2 2 1 0] - I can select the first 2 out of 5 in 5C2=10 ways - for the second 2 I can select 2 out of 3 3C2= 3 way Multiplying 10*3=30

What is wrong with this approach?

This is fine but you missed the last step. You have selected "first two" and "second two" i.e. you have arranged them in first and second. You need to un-arrange them. So you have to divide by 2. You get 30/2 = 15
_________________

In how many ways can 4 different marbles be distributed in 5 identical pockets?

Now, in this scenario do we have to select pockets first, which is 5C4 and then put marbles in 4! ?

Any thoughts.

Regards, Gaurav

If the pockets are identical, how would you select 4 out of the 5? You can pick 4 of the 5 identical pockets in just 1 way. You have to decide how to distribute marbles and then find the number of ways you can do that.

(4, 0, 0, 0, 0) - All 4 marbles in 1 pocket in 1 way. All pockets are identical. (3, 1, 0, 0, 0) - Select 1 of the 4 marbles in 4C1 ways = 4 ways and put it separately in one pocket. Put other 3 marbles in any other pocket in 1 way. So total 4 ways. (2, 2, 0, 0, 0) - We have to split 4 marbles in 2 groups of 2 marbles each. This can be done in 4!/(2!*2!*2!) = 3 ways (2, 1, 1, 0, 0) - Select 2 of the 4 marbles in 4C2 = 6 ways and put them in one pocket. Put the other 2 marbles in other 2 pockets in 1 way. Total number of ways = 6 (1, 1, 1, 1, 0) - Put the 4 marbles individually in any 4 pockets in 1 way.

This gives us a total of 1 + 4 + 3 + 6 + 1 = 15 ways

Check out this post for a similar question (Question 3):

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

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22 Mar 2016, 04:12

VeritasPrepKarishma wrote:

gmatminator wrote:

VeritasPrepKarishma wrote:

As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to un-arrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15

If you present your calculation, I can tell you what you are missing.

I am getting 30 as well, here is my reasoning: [2 2 1 0] - I can select the first 2 out of 5 in 5C2=10 ways - for the second 2 I can select 2 out of 3 3C2= 3 way Multiplying 10*3=30

What is wrong with this approach?

This is fine but you missed the last step. You have selected "first two" and "second two" i.e. you have arranged them in first and second. You need to un-arrange them. So you have to divide by 2. You get 30/2 = 15

If we need to un-arrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's un-arrange them => 10/2! = 5

Please, tell me what point did I miss.
_________________

I’m not afraid of the man who knows 10,000 kicks and has practiced them once. I am afraid of the man who knows one kick & has practiced it 10,000 times! - Bruce Lee

Please, press the +1 KUDOS button , if you find this post helpful

If we need to un-arrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's un-arrange them => 10/2! = 5

Please, tell me what point did I miss.

Putting 3 in a basket is different from putting 2 in a basket so we select 3 out of 5 and put those in one basket. The rest of the 2 we put in another basket and that's it. You are selecting 3 the first time and have 2 leftover.

In case we are splitting 4 in 2 groups of 2, we get two identical cases. You are selecting 2 the first time and have 2 leftover. So, putting A and B in a basket and C and D in another basket is the same as putting C and D in one basket and A and B in another basket. Hence we divide by 2! here.
_________________

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

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23 Mar 2016, 08:50

VeritasPrepKarishma wrote:

leeto wrote:

If we need to un-arrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's un-arrange them => 10/2! = 5

Please, tell me what point did I miss.

Putting 3 in a basket is different from putting 2 in a basket so we select 3 out of 5 and put those in one basket. The rest of the 2 we put in another basket and that's it. You are selecting 3 the first time and have 2 leftover.

In case we are splitting 4 in 2 groups of 2, we get two identical cases. You are selecting 2 the first time and have 2 leftover. So, putting A and B in a basket and C and D in another basket is the same as putting C and D in one basket and A and B in another basket. Hence we divide by 2! here.

"In case we are splitting 4 in 2 groups of 2, we get two identical cases." To make thinks easier to member and eventually digest, could I assume that this is some kind of the "Mississippi Problem" logic ? So, for example, let say the same problem only - we have 6 fruits and 4 baskets and in case {3,3, 0, 0 } - I will divide all out expression on (2!) because we get two identical cases => I mean the whole expression probably will be (6C3*3C3) / (2!) = 10 And, in the same problem with 10 fruits and 4 baskets, in the case {3,3,2,2} - We'll divide all expression on (2!)*(2!) ( because two identical cases for 3 fruits and two identical cases for 2 fruits - So it will be look like (10C3*7C3*4C2*2C2) / (2! * 2! ) = 60 * 35 * 3 = 60*105 = 6300 Do I correctly understand or I miss something again ? P.S. Excuses me for my meticulousness and many thanks for your helps.
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I’m not afraid of the man who knows 10,000 kicks and has practiced them once. I am afraid of the man who knows one kick & has practiced it 10,000 times! - Bruce Lee

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Last edited by leeto on 29 Mar 2016, 00:16, edited 1 time in total.

If we need to un-arrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's un-arrange them => 10/2! = 5

Please, tell me what point did I miss.

Putting 3 in a basket is different from putting 2 in a basket so we select 3 out of 5 and put those in one basket. The rest of the 2 we put in another basket and that's it. You are selecting 3 the first time and have 2 leftover.

In case we are splitting 4 in 2 groups of 2, we get two identical cases. You are selecting 2 the first time and have 2 leftover. So, putting A and B in a basket and C and D in another basket is the same as putting C and D in one basket and A and B in another basket. Hence we divide by 2! here.

"In case we are splitting 4 in 2 groups of 2, we get two identical cases." To make thinks easier to member and eventually digest, could I assume that this is some kind of the "Mississippi Problem" logic ? So, for example, let say the same problem only - we have 6 fruits and 4 baskets and in case {3,3, 0, 0 } - I will divide all out expression on (2!) because we get two identical cases => I mean the whole expression probably will be (6C3*3C3) / (2!) = 10 And, in the same problem with 10 fruits and 4 baskets, in the case {3,3,2,2} - We'll divide all expression on (2!)*(2!) ( because two identical cases for 3 fruits and two identical cases for 2 fruits - So it will be look like (10C3*7C3*4C2*2C2) / (2! * 2! ) = 60 * 35 * 3 = 60*105 = 6300 Do I correctly understand or I miss something again ? P.S. Excuses me for my meticulousness and many thanks for your helps.

Yes, you are right. You have understood the concept.
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In how many ways can 5 different marbles be distributed in 4 identical pockets?

A. 24 B. 51 C. 120 D. 625 E. 1024

The marbles only need to be distinguished and so the problem boils down to selection of marbles grouped in different ways.

There are 6 different types of grouping possible.

Case 1: 5 marbles together. The 5 marbles can be selected in 1 way. Case 2: 4 marbles together and 1 marble separately. The marbles can be selected in 5C4 ways= 5 ways Case 3: 3 marbles together and 2 marbles together. The marbles can be selected in 5C3 ways=10 ways Case 4: 3 marbles together and two 1 marble each. The marbles can be selected in 5C3 ways=10 ways Case 5: two set of 2 marbles and 1 marble. The marbles can be selected in 5C2*3C2/2! ways = 15 ways case 6: 2 marbles together and three 1 marble. The marbles can be selected in 5C2 ways= 10 ways

Total 51

In the case of 5, to clarify see that (1,2) (3,4) 5 and (3,4) (1,2) and 5 are the same and so on. This is because we are selecting 2 marbles twice i.e., first 5C2 and then 3C2.. So we divide by 2!.
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