Combinations

[Prajakt]

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

[GMATGuruNY]

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.

[Prajakt]

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

[vk_vinayak]

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]

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

Cheers,
Brent

[mgm]

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 ?

[Brent@GMATPrepNow]

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 ?

I'd be surprised if there were these kinds of distribution questions on the GMAT.

That said, if there were such a question, I think that only 800-level test-takers would encounter it.

Of course, some of the strategies/concepts (not necessarily the formula though) used in the above solutions might prove useful for other counting questions.

Cheers,
Brent

[[email protected]]

Hi Prajakt,

This question is badly worded (and as such would NOT appear on the GMAT as is). The first sentence tells us that the chocolates are going to be divided among 3 PEOPLE (who I would assume to be unique). The next sentence THEN says the chocolates are distributed into PARTS. 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. Here's why...

If our 3 groups are 10, 0 and 0 (in any order) then it doesn't matter which one is the "10". This represents 1 option.

However If our 3 groups are 10, 0 and 0 AND these groups are given to 3 different people (I'll call them A, B, and C), then there are 3 options
A = 10, B = 0, C = 0
A = 0, B = 10, C = 0
A = 0, B = 0, C = 10

This would echo throughout the rest of the calculations.

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

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.

[[email protected]]

Hey Mitch,

You are correct. As I reviewed my work on this question, I realize I had mis-organized my work.

GMAT assassins aren't born, they're made,
Rich