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oranges

by goyalsau » Thu Dec 02, 2010 7:57 am
If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30 b. 210 c. 15 d. 35 e. 40
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by shovan85 » Thu Dec 02, 2010 8:22 am
goyalsau wrote:If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30 b. 210 c. 15 d. 35 e. 40
There is a formula in Combinatorics:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0, 1, 2 or more items is C(n+r-1, r-1)
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is C(n-1,r-1)

PS: I have no proof of this. Try with a small number (2,3) you can visualize this:
Say 2 oranges are to be divided among 2 children (no at least, One can have all)
Then possibilities are {A,A},{AA,0},{0,AA}. So, 3 ways.
Formula C(2+2-1,2-1) = C(3,1) = 3
Come to problem:

All 4 children get at least 3 each
Thus, 4*3 = 12 is the least number of Oranges distributed among the 4.

Now Remaining 16 - 12 = 4 Oranges are to be distributed among 4 children.

Apply the formula, The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0, 1, 2 or more items is C(n+r-1, r-1)

n = 4, r = 4

Thus C(4+4-1,4-1) = C(7,3) = 35

Pick D
Last edited by shovan85 on Thu Dec 02, 2010 9:10 am, edited 1 time in total.
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by Rahul@gurome » Thu Dec 02, 2010 8:54 am
goyalsau wrote:If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30 b. 210 c. 15 d. 35 e. 40
If each of them gets at least 3 oranges, give each of them 3 oranges. Number of oranges left = (16 - 3*4) = 4. The question now can be simply rephrased as "In how many ways 4 oranges can be distributed among 4 children?" Which in turn equivalently it can be rephrased as "In how many ways 4 oranges can be separated into four groups so that a group may not contain an orange?"

Apply the famous "Separator" method.

Say, each orange is represented by '*' and separator as '|'.
Some of possible grouping and their representations are,
  • 1, 1, 0, 2 --> *|*||**
    3, 0, 0, 1 --> ***|||*
    0, 4, 0, 0 --> |****||
Number of grouping simply becomes number of ways to arrange 4 oranges and 3 separators between themselves = (4 + 3)!/(4!*3!) = 35

The correct answer is D.

Proof of Shovan's formula no. 1:
If we had to distribute n identical items among r persons such that some of them may not get an item,
  • Number of separators = (r - 1)
    Number of ways to distribute = Number of ways to arrange n items and (r - 1) separators between themselves = (n + r -1)!/(n!*(r - 1)!) = (n + r - 1)C(r - 1)
Proof of Shovan's formula no. 2:
See this link: https://www.beatthegmat.com/12-t70715.html#319654
It is easy to get the proof from the above link.
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by goyalsau » Thu Dec 02, 2010 9:24 am
Rahul@gurome wrote: Apply the famous "Separator" method.

Say, each orange is represented by '*' and separator as '|'.
Some of possible grouping and their representations are,
  • 1, 1, 0, 2 --> *|*||**
    3, 0, 0, 1 --> ***|||*
    0, 4, 0, 0 --> |****||
Number of grouping simply becomes number of ways to arrange 4 oranges and 3 separators between themselves = (4 + 3)!/(4!*3!) = 35
Rahul you have not subtracted 2 from 35 because it was a single digit number 4

Lets says if they have to distribute 18 oranges in 4 people { after the first 4 oranges }

then it would be (18 + 3 ) ! / 18! * 3 ! = 1330 ways,
1330 - 2 = 1328

Please correct me if am wrong, What if the there are 19 oranges because then we will require at least 3 single digits to fill,
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by Rahul@gurome » Thu Dec 02, 2010 9:34 am
goyalsau wrote:Rahul you have not subtracted 2 from 35 because it was a single digit number 4

Lets says if they have to distribute 18 oranges in 4 people { after the first 4 oranges }

then it would be (18 + 3 ) ! / 18! * 3 ! = 1330 ways,
1330 - 2 = 1328

Please correct me if am wrong, What if the there are 19 oranges because then we will require at least 3 single digits to fill,
In case of numbers with digits (the link of the post I mentioned) we had an upper limit of values because a digit can't have a value greater than 9. But in case of oranges or any items we don't have such an upper limit or any constraints like each of them will get at most some oranges. Thus we don't need to subtract them.

Hope it helps.
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by justvishu4 » Thu Dec 02, 2010 12:35 pm
Rahul@gurome wrote:
goyalsau wrote: Number of grouping simply becomes number of ways to arrange 4 oranges and 3 separators between themselves = (4 + 3)!/(4!*3!) = 35
Rahul - quick (dumb?) question-

How did you arrive that there are three separators?
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by Rahul@gurome » Thu Dec 02, 2010 12:52 pm
justvishu4 wrote:How did you arrive that there are three separators?
We have to distribute the oranges in four groups.
Three separators needed to divide them in four groups because there is three separation (or gaps) between four groups. It's like placing the oranges in a row and putting three chalk mark between them to divide them into four groups.

Say we have 10 oranges ( represented by *)
Grouping can be done as: ***|*|***|***

In general, if we have to distribute into r groups, (r - 1) separators are needed.

Hope that helps.
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by justvishu4 » Thu Dec 02, 2010 1:32 pm
Rahul@gurome wrote:
justvishu4 wrote:How did you arrive that there are three separators?
We have to distribute the oranges in four groups.
Three separators needed to divide them in four groups because there is three separation (or gaps) between four groups. It's like placing the oranges in a row and putting three chalk mark between them to divide them into four groups.

Say we have 10 oranges ( represented by *)
Grouping can be done as: ***|*|***|***

In general, if we have to distribute into r groups, (r - 1) separators are needed.

Hope that helps.
Yup, makes sense. Thanks Rahul.
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by goyalsau » Thu Dec 02, 2010 9:42 pm
shovan85 wrote:
There is a formula in Combinatorics:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0, 1, 2 or more items is C(n+r-1, r-1)
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is C(n-1,r-1)
Say 2 oranges are to be divided among 2 children (no at least, One can have all)
Then possibilities are {A,A},{AA,0},{0,AA}. So, 3 ways.
Formula C(2+2-1,2-1) = C(3,1) = 3
Thanks Shovan , Awesome work, You have given the formula and Rahul has given the proof,
Thanks Rahul, I Really don't know how to thank you for all the Help, But i must say one thing.

If Nobel prize Jury ask me for a opinion I will recommend your Name.
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by shovan85 » Thu Dec 02, 2010 10:49 pm
goyalsau wrote: Thanks Rahul, I Really don't know how to thank you for all the Help, But i must say one thing.

If Nobel prize Jury ask me for a opinion I will recommend your Name.
Dude!! sadly there is no Nobel Prize awarded for mathematics. We will refer Rahul's name for Fields Medal :)
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by arora007 » Mon Jan 31, 2011 1:52 am
Rahul@gurome wrote:
goyalsau wrote:If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30 b. 210 c. 15 d. 35 e. 40
If each of them gets at least 3 oranges, give each of them 3 oranges. Number of oranges left = (16 - 3*4) = 4. The question now can be simply rephrased as "In how many ways 4 oranges can be distributed among 4 children?" Which in turn equivalently it can be rephrased as "In how many ways 4 oranges can be separated into four groups so that a group may not contain an orange?"

Apply the famous "Separator" method.

Say, each orange is represented by '*' and separator as '|'.
Some of possible grouping and their representations are,
  • 1, 1, 0, 2 --> *|*||**
    3, 0, 0, 1 --> ***|||*
    0, 4, 0, 0 --> |****||
Number of grouping simply becomes number of ways to arrange 4 oranges and 3 separators between themselves = (4 + 3)!/(4!*3!) = 35

The correct answer is D.

Proof of Shovan's formula no. 1:
If we had to distribute n identical items among r persons such that some of them may not get an item,
  • Number of separators = (r - 1)
    Number of ways to distribute = Number of ways to arrange n items and (r - 1) separators between themselves = (n + r -1)!/(n!*(r - 1)!) = (n + r - 1)C(r - 1)
Proof of Shovan's formula no. 2:
See this link: https://www.beatthegmat.com/12-t70715.html#319654
It is easy to get the proof from the above link.
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