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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Arabian Horses  Good One!
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Take the task of assigning the 4 carts break it into stages.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
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/gmatcounting?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/gmatcounting?id=775

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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
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
Thanks Brent for the prompt and detailed answer.[email protected] wrote:Take the task of assigning the 4 carts break it into stages.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
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/gmatcounting?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/gmatcounting?id=775
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.
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I think this is a 700+ level question, but I don't think it's out of scope (IMHO)puneetkhurana2000 wrote:Note: This question should be out of scope for GMAT.
Cheers,
Brent
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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.tabsang wrote:Thanks Brent for the prompt and detailed answer.[email protected] wrote:Take the task of assigning the 4 carts break it into stages.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
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/gmatcounting?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/gmatcounting?id=775
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.
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!!![email protected] wrote: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.tabsang wrote:Thanks Brent for the prompt and detailed answer.[email protected] wrote:Take the task of assigning the 4 carts break it into stages.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
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/gmatcounting?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/gmatcounting?id=775
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.
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
(I'm so happy that the nonformula 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 nonformula driven approach to this and it'll make my day
Cheers,
Taz
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I suggest that you first pretend that the n groups are distinct.tabsang wrote:
Awesome!! Fantastic!! Thanks Brent, you rock!!!
(I'm so happy that the nonformula 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 nonformula driven approach to this and it'll make my day
Cheers,
Taz
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.[email protected] wrote:I suggest that you first pretend that the n groups are distinct.tabsang wrote:
Awesome!! Fantastic!! Thanks Brent, you rock!!!
(I'm so happy that the nonformula 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 nonformula driven approach to this and it'll make my day
Cheers,
Taz
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
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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[email protected] wrote:Take the task of assigning the 4 carts break it into stages.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
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/gmatcounting?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/gmatcounting?id=775
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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Yes, that is correct.muhtasimhassan wrote:[email protected] wrote:Take the task of assigning the 4 carts break it into stages.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
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/gmatcounting?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/gmatcounting?id=775
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
Cheers,
Brent
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.
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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That's a lot more complicated! We'd need to know whether the groups are distinct, but assuming that they're not: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?
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!