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

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Question Stats: 47% (01:56) correct 53% (02:01) wrong based on 76 sessions

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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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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

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

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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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1
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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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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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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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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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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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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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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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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Posts: 8187
Re: In how many ways, 10 identical chocolates be distributed among 3 child  [#permalink]

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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.

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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