Problem Solving

[email protected] wrote:

tabsang wrote:

[email protected] wrote:

tabsang wrote:Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible?

a) 420
b) 1260
c) 5220
d) 9450
e) 113400

OA is E

Take the task of assigning the 4 carts break it into stages.

Stage 1: Assign the first cart to 2 horses
There are 10 horses and we must select 2.
Since the order of the selected horses does not matter, we can use combinations.
We can select 2 horses from 10 horses in 10C2 ways (45 ways).

Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Assign the second cart to 2 horses
There are 8 horses remaining and we must select 2.
We can select 2 horses from 8 horses in 8C2 ways (28 ways).

Stage 3: Assign the third cart to 2 horses
There are 6 horses remaining and we must select 2.
We can select 2 horses from 6 horses in 6C2 ways (15 ways).

Stage 4: Assign the fourth cart to 2 horses
There are 4 horses remaining and we must select 2.
We can select 2 horses from 8 horses in 4C2 ways (6 ways).

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways ([spoiler]= 113,400 ways[/spoiler])

Answer = E

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

Thanks Brent for the prompt and detailed answer.

However, there's another problem.
I had, infact, solved the problem using the same approach as the one you took as opposed to the formula driven one. But this approach seemingly fails if I try and apply it to this question:

In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

a) 280
b) 1,260
c) 1,680
d) 2,520
e) 3,360

OA is a.
If I apply the same logic as that I used in the Arabian horses problem then the answer comes out to be c .

Please help. I'm confused.

If the 3 groups are distinct (e.g., being in group A is different from being in group B) then the correct answer to this question is, indeed, C. However, if the 3 groups are not distinct, then the correct answer is A.

From the wording of the question, it's difficult to determine whether the 3 groups are distinct. This ambiguity makes the question unanswerable.

Cheers,
Brent

[email protected] wrote:

tabsang wrote:
Awesome!! Fantastic!! Thanks Brent, you rock!!!
(I'm so happy that the non-formula approach holds true, albeit only if the groups are distinct)

Brent, I have another quick question to bring a full closure to this problem and concept:
In case the groups were not distinct, what would be the approach to solve the problem?

I know of a formula that we can use when the order is not important or as in this case the groups are not distinct (for (mxn) things distributed equally in groups of m each containing n objects and the order of the groups is not important: (mn)!/[(n!^m)*m!])

Please tell me that there is a non-formula driven approach to this and it'll make my day

Cheers,
Taz

I suggest that you first pretend that the n groups are distinct.
Then, to handle the fact that they are not distinct, divide your earlier result by n!
We divide by n! because we can take n things (groups) and arrange them in n! ways.

So, for the question about 9 people, we'll first assume that the groups are distinct.
When we do so, we see that there are 1680 different possibilities.

Then, to account for the fact that the 3 groups are not distinct, divide 1680 by 3! to get 280

Cheers,
Brent

[email protected] wrote:

tabsang wrote:Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible?

a) 420
b) 1260
c) 5220
d) 9450
e) 113400

OA is E

Take the task of assigning the 4 carts break it into stages.

Stage 1: Assign the first cart to 2 horses
There are 10 horses and we must select 2.
Since the order of the selected horses does not matter, we can use combinations.
We can select 2 horses from 10 horses in 10C2 ways (45 ways).

Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Assign the second cart to 2 horses
There are 8 horses remaining and we must select 2.
We can select 2 horses from 8 horses in 8C2 ways (28 ways).

Stage 3: Assign the third cart to 2 horses
There are 6 horses remaining and we must select 2.
We can select 2 horses from 6 horses in 6C2 ways (15 ways).

Stage 4: Assign the fourth cart to 2 horses
There are 4 horses remaining and we must select 2.
We can select 2 horses from 8 horses in 4C2 ways (6 ways).

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways ([spoiler]= 113,400 ways[/spoiler])

Answer = E

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775