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## Arabian Horses - Good One!

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### Arabian Horses - Good One!

by tabsang » Wed Dec 12, 2012 11:05 am
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

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by [email protected] » Wed Dec 12, 2012 11:15 am
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])

Cheers,
Brent

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by puneetkhurana2000 » Wed Dec 12, 2012 11:21 am
Note:- This question should be out of scope for GMAT.

The solution is: -
Number of ways 10 horses can be divided into 5 groups when order of the groups does not matter is:

(10C2*8C2*6C2*4C2*2C2)/5! = 945

Now to assign a pair to one of 4 distinct is 5P4 = 120.

Total is 945 * 120 = 113400

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by tabsang » Wed Dec 12, 2012 11:46 am
Last edited by tabsang on Wed Dec 12, 2012 11:52 am, edited 1 time in total.

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by tabsang » Wed Dec 12, 2012 11:50 am
[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])

Cheers,
Brent

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 .

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by [email protected] » Wed Dec 12, 2012 12:14 pm
puneetkhurana2000 wrote:Note:- This question should be out of scope for GMAT.
I think this is a 700+ level question, but I don't think it's out of scope (IMHO)

Cheers,
Brent
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by [email protected] » Wed Dec 12, 2012 12:18 pm
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])

Cheers,
Brent

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 .

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
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by tabsang » Wed Dec 12, 2012 12:28 pm
[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])

Cheers,
Brent

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 .

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
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

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by [email protected] » Wed Dec 12, 2012 12:58 pm
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
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by tabsang » Wed Dec 12, 2012 1:06 pm
[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
Thanks Brent.
You, sir, are a life saver
I had been at this for quite some time and now it rests in peace with all concepts understood and learnt.

Have a great day

Cheers,
Taz

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by muhtasimhassan » Mon Jan 26, 2015 10:50 pm
[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])

Cheers,
Brent

Hi Brent,

This is a great explanation.

I have a question here. If it were mentioned in the question that the carts are not 'distinct', would we have divided (45)(28)(15)(6) by 4! ?

Thanks,
Muhtasim Hassan

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by [email protected] » Tue Jan 27, 2015 7:09 am
muhtasimhassan 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])

Cheers,
Brent

Hi Brent,

This is a great explanation.

I have a question here. If it were mentioned in the question that the carts are not 'distinct', would we have divided (45)(28)(15)(6) by 4! ?

Thanks,
Muhtasim Hassan
Yes, that is correct.

Cheers,
Brent
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by Kaustubhk » Tue May 23, 2017 8:27 am
Hi Brent,

If the question is rephrased as below

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

How we will solve this?

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by Kaustubhk » Wed May 24, 2017 8:13 am
Hi Brent,

144
234
126
333
522
711
531

144 can arranged in 3!/2! = 3 ways
234 can arranged in 3! = 6 ways
126 can arranged in 3! = 6 ways
333 can arranged in 1! = 1 way
522 can arranged in 3!/2! = 3 ways
711 can arranged in 3!/2! = 3 ways
531 can arranged in 3! = 6 ways

Total number of ways = 28 ways.

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by [email protected] » Wed May 24, 2017 4:33 pm
Kaustubhk wrote:Hi Brent,

If the question is rephrased as below

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

How we will solve this?
That's a lot more complicated! We'd need to know whether the groups are distinct, but assuming that they're not:

First, choose 3 people for the first group: (9 choose 3)

Next, choose 3 people for the second group: (6 choose 3)

The three people remaining form the third group, so we're almost done. The only step left is to remove all the duplicates. For instance, we could have Al, Bill, and Carl as group 1, but we could also have Al, Bill, and Carl as group 2. To eliminate the duplicate arrangements, we divide by all the ways we could arrange the groups once they're formed: 3 * 2 * 1.

That leaves us with (9 choose 3) * (6 choose 3) / 3!