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In how many ways, 10 identical chocolates be distributed among 3 child

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In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post Updated on: 07 Oct 2019, 00:27
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In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.

Originally posted by Kjol on 06 Oct 2019, 22:06.
Last edited by Kjol on 07 Oct 2019, 00:27, edited 1 time in total.
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 07 Oct 2019, 09:44
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Top Contributor
Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.


Another approach is to look for a pattern

Say the 3 children are A, B, C
We'll denote each outcome as follows: # chocolates for A | # chocolates for B | # chocolates for C

Number of outcomes in which child A receives exactly 1 chocolate
1 | 1 | 8
1 | 2 | 7
1 | 3 | 6
1 | 4 | 5
1 | 5 | 4
1 | 6 | 3
1 | 7 | 2
1 | 8 | 1
8 outcomes


Number of outcomes in which child A receives exactly 2 chocolates
2 | 1 | 7
2 | 2 | 6
2 | 3 | 5
2 | 4 | 4
2 | 5 | 3
2 | 6 | 2
2 | 7 | 1
7 outcomes


Number of outcomes in which child A receives exactly 3 chocolates
3 | 1 | 6
3 | 2 | 5
3 | 3 | 4
3 | 4 | 3
3 | 5 | 2
3 | 6 | 1
6 outcomes

See the pattern?

So, the TOTAL number of outcomes = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= 36

Answer: A

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In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 07 Oct 2019, 00:12
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Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.


Two ways..

Formula:-
First give 1 each to the three children, so we are left with 10-3=7.
Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)

first give one to each, hence 3 gone. Now you can distribute the remaining 7 to anyone as the restriction of at least one is done..
A+B+C=7....
Now, take this as 7 marbles and 2 partitions, so total 9 things. You can choose these 2 partitions in 9C2 ways=\(\frac{9!}{7!2!}=\frac{9*8}{2}=36\)

So, the formula \((n+k-1)C(k-1)\) comes from k items, so k-1 partition and then choosing these k-1 partition from it..
Here n=7, and k=3..thus \((n+k-1)C(k-1)=(7+3-1)C(3-1)=9C2\)


Calculations:-

Let the number each can get be-
8,1,1-3!/2=3 ways
7,2,1--3! ways
6,3,1--3!

6,2,2 -- 3!/2=3
5,4,1 -- 3!
5,3,2 -- 3!

4,4,2 --3!/2=3
4,3,3 -- 3!/2=3
Total = 4*3!+4*3=4*6+12=36

A
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 07 Oct 2019, 00:14
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The possible combinations are:
118 --> with 3 rearrangements ( because 1 is repeated, the possibilities = \(\frac{3!}{2!} = 3\))
127 --> with 6 rearrangements (because all numbers are different, the possibilities = \(3! = 6\))
136 --> with 6 rearrangements
145 --> with 6 rearrangements
226 --> with 3 rearrangements
235 --> with 6 rearrangements
244 --> with 3 rearrangements
334 --> with 3 rearrangements

Total = 36

Another faster way is the "Stars and Bars Theorem", which says that for n items distributed on k elements, where each of the elements must have a positive value, then the number of possibilities = (n-1)C(k-1).
and in our case, it will be (10-1)C(3-1) = 9C2 = 36
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 07 Oct 2019, 09:10
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Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.



After distributing 1 chocolate to each child
Remaining 7 chocolates are to be distributed among 3 children

*|*|*

This can be done in (7+2)!/7!2! = 9!/7!2! = 9*8/2 = 36

IMO A
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 07 Oct 2019, 09:39
Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.


Formula: Number of positive integral solutions of the equation \(x_1\) + \(x_2\) + . . . . + \(x_r\) = n is \((n-1)c_{r-1}\)

We have to find \(x_1\) + \(x_2\) + \(x_3\) = 10
—> Number of positive integral solutions = \((10-1)c_{3-1}\) = \(9c_2\) = 9*8/2! = 36

IMO Option A

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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 07 Oct 2019, 19:24
chetan2u wrote:
Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.


Two ways..

Formula:-
First give 1 each to the three children, so we are left with 10-3=7.
Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)

Calculations:-

Let the number each can get be-
8,1,1-3!/2=3 ways
7,2,1--3! ways
6,3,1--3!

6,2,2 -- 3!/2=3
5,4,1 -- 3!
5,3,2 -- 3!

4,4,2 --3!/2=3
4,3,3 -- 3!/2=3
Total = 4*3!+4*3=4*6+12=36

A



Hello Chetan2u,

Whats this formula approach? Can you please explain ?


I was able to understand the calculation approach.
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In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 11 Oct 2019, 06:25
1
Another Way to solve it is to make a list where you start each point of the list with how many chocolates you give to the first child, so the list is
8
7
6 usw.

Then you check in how many ways you can distribute the remaining chocolates among the other two children.
For 8 it is 1
For 7 It is 2 usw, in the end you add all the numbers between 1 and 8 inclusive, and the sum is 36
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 13 Oct 2019, 01:31
chetan2u wrote:
Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.


Two ways..

Formula:-
First give 1 each to the three children, so we are left with 10-3=7.
Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)

Calculations:-

Let the number each can get be-
8,1,1-3!/2=3 ways
7,2,1--3! ways
6,3,1--3!

6,2,2 -- 3!/2=3
5,4,1 -- 3!
5,3,2 -- 3!

4,4,2 --3!/2=3
4,3,3 -- 3!/2=3
Total = 4*3!+4*3=4*6+12=36

A



Could you please explain why you're doing 9C7 instead of 10C7 after distributing one to each child?
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 17 Oct 2019, 14:05
chetan2u Can you please explain the formula approach
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 18 Oct 2019, 09:51
chetan2u wrote:
Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.


Two ways..

Formula:-
First give 1 each to the three children, so we are left with 10-3=7.
Now you can distribute these 7 to any child in \((n-1)C7=9C7=\frac{9!}{2!7!}=36\)

Calculations:-

Let the number each can get be-
8,1,1-3!/2=3 ways
7,2,1--3! ways
6,3,1--3!

6,2,2 -- 3!/2=3
5,4,1 -- 3!
5,3,2 -- 3!

4,4,2 --3!/2=3
4,3,3 -- 3!/2=3
Total = 4*3!+4*3=4*6+12=36

A


chetan2u

why do we have to do "(n−1)" ?
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Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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New post 26 Oct 2019, 21:34
1
Kjol wrote:
In how many ways, 10 identical chocolates be distributed among 3 children such that each child gets at least 1 chocolate?

A. 36
B. 66
C. 72
D. 78
E. 84.



first give one to each, hence 3 gone. Now you can distribute the remaining 7 to anyone as the restriction of at least one is done..
A+B+C=7....
Now, take this as 7 marbles and 2 partitions, so total 9 things. You can choose these 2 partitions in 9C2 ways=\(\frac{9!}{7!2!}=\frac{9*8}{2}=36\)

So, the formula (n+k-1)C(k-1) comes from k items, so k-1 partition and then choosing these k-1 partition from it..
Here n=7, and k=3..thus (n+k-1)C(k-1)=(7+3-1)C(3-1)=9C2

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Re: In how many ways, 10 identical chocolates be distributed among 3 child   [#permalink] 26 Oct 2019, 21:34
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