Problem Solving

Suppose I have 10 identical chocolates to be divided among 3 people. These 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates. In how many ways this can be done?
A. 120
B. 66
C. 132
D. 1210
E. 75

Prajakt wrote:Suppose I have 10 identical chocolates to be divided among 3 people. These 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates. In how many ways this can be done?
A. 120
B. 66
C. 132
D. 1210
E. 75

The following is called the SEPARATOR method.

Ten identical chocolates are to be separated into -- at most -- 3 groupings.
Thus, we need ten chocolates and two separators:
OOOO|OOO|OOO

Each arrangement of the elements above represents one way to distribute the ten chocolates among three people A, B and C:
OOOO|OOO|OOO = A gets 4 chocolates, B gets 3 chocolates, C gets 3 chocolates.
OO||OOOOOOOO = A gets 2 chocolates, B gets 0 chocolates, C gets 8 chocolates.
OOOOOOOOOO|| = A gets all 10 chocolates.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 12 elements above (the 10 identical chocolates and the two identical separators).
The number of ways to arrange 12 elements = 12!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, the number of ways to arrange the 10 identical chocolates and the 2 identical separators is equal to the following:
12!/10!2! = 66.

The correct answer is B.

GMATGuruNY wrote:

Prajakt wrote:Suppose I have 10 identical chocolates to be divided among 3 people. These 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates. In how many ways this can be done?
A. 120
B. 66
C. 132
D. 1210
E. 75

The following is called the SEPARATOR method.

Ten identical chocolates are to be separated into -- at most -- 3 groupings.
Thus, we need ten chocolates and two separators:
OOOO|OOO|OOO

Each arrangement of the elements above represents one way to distribute the ten chocolates among three people A, B and C:
OOOO|OOO|OOO = A gets 4 chocolates, B gets 3 chocolates, C gets 3 chocolates.
OO||OOOOOOOO = A gets 2 chocolates, B gets 0 chocolates, C gets 8 chocolates.
OOOOOOOOOO|| = A gets all 10 chocolates.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 12 elements above (the 10 identical chocolates and the two identical separators).
The number of ways to arrange 12 elements = 12!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, the number of ways to arrange the 10 identical chocolates and the 2 identical separators is equal to the following:
12!/10!2! = 66.

The correct answer is B.

I use below mentioned formula whenever N identical items have to be divided into R groups such that each group can get some or none.
N+R-1 C R-1

Prajakt wrote:I use below mentioned formula whenever N identical items have to be divided into R groups such that each group can get some or none.
N+R-1 C R-1

Even I used the formula to solve. Going by Mitch's post, it seems that SEPARATOR method is easier to understand and apply than remember the formula.

Brent@GMATPrepNow wrote:If anyone is interested, here's a related question to try: https://www.beatthegmat.com/very-tricky- ... 25349.html

Cheers,
Brent

Brent are the distribution problems modeled after real GMAT problems ?

[email protected] wrote:If we're dividing into "parts" (or "groups"), then Mitch's solution is correct because we'd be looking for combinations of chocolates. However, this question states that the chocolates are divided among 3 people, and each person is unique. This changes the math/solution.

Not so.
The separator method used in my solution above takes into account that the 3 people are DISTINCT.

Here is how the separator method accounts for all of the ways to give 1 person 4 chocolates, while the other 2 people each receive 3 chocolates:
OOOO|OOO|OOO = A gets 4 chocolates, B gets 3 chocolates, C gets 3 chocolates.
OOO|OOOO|OOO = A gets 3 chocolates, B gets 4 chocolates, C gets 3 chocolates.
OOO|OOO|OOOO = A gets 3 chocolates, B gets 3 chocolates, C gets 4 chocolates.
All 3 ways are counted.

Here is how the separator method accounts for all of the ways to give 1 person 10 chocolates, while the other 2 people each receive 0 chocolates:
OOOOOOOOOO|| = A gets 10 chocolates, B gets 0 chocolates, C gets 0 chocolates.
|OOOOOOOOOO| = A gets 0 chocolates, B gets 10 chocolates, C gets 0 chocolates.
||OOOOOOOOOO = A gets 0 chocolates, B gets 0 chocolates, C gets 10 chocolates.
All 3 ways are counted.

In fact, if the order of the distributions DOESN'T MATTER -- if A=10, B=0 and C=0 is considered the same distribution as A=0, B=10, and C=0 -- then the separator method cannot be used, making the solution far more time-consuming.