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

Question

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.

Bunuel · Accepted Answer

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

When distributing 10 identical chocolates among 3 children, we can use "Stars and Bars" method to solve the problem.

Imagine the 10 chocolates as 10 stars in a row. To divide these stars among 3 children, we need 2 bars to create 3 separate sections. For example:

***|****|*** would represent that the first child got 3 chocolates, the second got 4, and the third got 3.
||********** would represent that the first child got 0 chocolates, the second got 0, and the third got 10.
****||****** would represent that the first child got 4 chocolates, the second got 0, and the third got 6.

So, the problem becomes one of arranging these 10 identical stars and 2 identical bars in a row, which is given by 12!/(10!2!) = 66.

Hope it helps.

Bunuel · Answer

The method of (number of groups)^(number of items to distribute) works if the items we distribute are distinct. For example, in how many ways can we distribute x, y, z between A and B? The answer is (number of groups)^(number of items to distribute) = 2^3 = 8.

--- A ----- B ---
x, y, z ----- 0
x, y ----- z
x, z ----- y
y, z ----- x
x ----- y, z
y ----- x, z
z ----- x, y
0 ----- x, y, z

However, if the items are not distinct, we should use the "Stars and Bars" method. For example, how many ways are there to distribute three identical items between A and B? The answer is 4!/3! = 4, which is the number of arrangements of ***|.

--- A ----- B ---
*** | 0
** | *
* | **
0 | ***

P.S. If, in the problem at hand, the chocolates were distinct, the question would become much harder, and your method would still be incorrect. The correct answer would be 3^10 minus the number of all arrangements where at least one child gets 0 chocolates.

P.P.S. Check this for more:  DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION)

chetan2u · Answer

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

MahmoudFawzy · Answer

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

BrentGMATPrepNow · Answer

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

Cheers, 
Brent

chetan2u · Answer

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

prgmatbiz   ManjariMishra   prabsahi  and  ShreyasJavahar

HAPPY DIWALI !!!!

kamalsharmam21 · Answer

What if the condition of at least is not there

Posted from my mobile device

kamalsharmam21 · Answer

Thanks awesome explanation 🙏

ahujaparth10 · Answer

BrentGMATPrepNow   Bunuel  Why dont we do 3^7 in this question? Each of 7 chocolates after distributing 1 chocolate can be given to any of the 3 children.

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

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