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

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

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

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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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

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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Hello Chetan2u,

Whats this formula approach? Can you please explain ?


I was able to understand the calculation approach.

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

Could you please explain why you're doing 9C7 instead of 10C7 after distributing one to each child?

chetan2u Can you please explain the formula approach

chetan2u

why do we have to do "(n−1)" ?

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 !!!!
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10 identical chocolates and 3 different children getting at least 1:
\(r=n:(x_1+1)+(x_2+1)+(x_3+1)=10\)
\(r=n:(x_1)+(x_2)+(x_3)=10-3=7\)
\(C(n+r-1;n,r-1)=\frac{9!}{7!2!}=36\)

Ans (A)

What if the condition of at least is not there

Posted from my mobile device

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.
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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.

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)
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