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combinatorics 3

by Gurpinder » Wed Jul 28, 2010 1:42 pm
A farmer has enough land to plant 3 crops. If there are 7 crops that will grow on the land, how many different groups of 3 different crops can be planted?

OA: 35

but shouldnt the answer be 210?

7! / 4! (because the three crops that are going to be planted are different as i interpret it from the bold part of the question).
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by selango » Wed Jul 28, 2010 2:00 pm
Its 7C3,not 7P3

7!/4!*3!=35

Whenever selecting one thing or so from a group,it is combination.
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by Brent@GMATPrepNow » Wed Jul 28, 2010 2:13 pm
Gurpinder wrote:A farmer has enough land to plant 3 crops. If there are 7 crops that will grow on the land, how many different groups of 3 different crops can be planted?

OA: 35

but shouldnt the answer be 210?

7! / 4! (because the three crops that are going to be planted are different as i interpret it from the bold part of the question).
This question is somewhat ambiguous.

If we assume that the order of the crops matters (e.g., corn on field 1, wheat on field 2 and flax on field 3 is different from wheat on field 1, flax on field 2 and corn on field 3), then your answer of 210 is correct (7x6x5=210)

If we assume that order does not matter (e.g., corn on field 1, wheat on field 2 and flax on field 3 is the same as wheat on field 1, flax on field 2 and corn on field 3), then this is a combination question and we are choosing 3 crops from 7. THis can be accomplished in 35 (aka 7C3) ways.
Brent Hanneson - Creator of GMATPrepNow.com
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