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

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

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.

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)

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 !!!!

What if the condition of at least is not there

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Thanks awesome explanation 🙏

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.