Problem Solving

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pmFollowed by: 6 members

by BTGmoderatorRO » Fri Nov 24, 2017 7:42 am

10 Arabian horses are split into pairs to pull one of the distinct 4 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

can someone explain further on why D is the answer. Pls an expert should come to my aid here

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Source: Beat The GMAT — Problem Solving | Fri Nov 24, 2017 7:42 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Nov 24, 2017 8:49 am

Roland2rule wrote:10 Arabian horses are split into pairs to pull one of the distinct 4 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

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

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html

MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html

DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8138; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Oct 13, 2019 5:09 pm

BTGmoderatorRO wrote:10 Arabian horses are split into pairs to pull one of the distinct 4 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

can someone explain further on why D is the answer. Pls an expert should come to my aid here

The first cart has 10C2 = (10 x 9)/2 = 45 ways to choose a pair of horses (or 2 horses).

The second cart has 8C2 = (8 x 7)/2 = 28 ways to choose a pair of horses.

The third cart has 6C2 = (6 x 5)/2 = 15 ways to choose a pair of horses.

Finally, the fourth, or last, cart has 4C2 = (4 x 3)/2 = 6 ways to choose a pair of horses.

Therefore, there are 45 x 28 x 15 x 6 = 113,400 ways to assign 10 horses to 4 carts.

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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