so, 12C4 = 12!/4! 8! = 9.10.11.12/1.2.3.4 = 9.5.11 = 45.11 = 495
Skywalker wrote:Why is the answer here option 3 i.e. 495. Please let me know,
GMAT Prep: Combination sum
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ashokkadam
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Since we just need combinations and order does not matter.
so, 12C4 = 12!/4! 8! = 9.10.11.12/1.2.3.4 = 9.5.11 = 45.11 = 495
so, 12C4 = 12!/4! 8! = 9.10.11.12/1.2.3.4 = 9.5.11 = 45.11 = 495
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Solution:
Since order does not matter here, it is a combination problem.
Number of ways of selecting 4 sites from given 12 sites is 12C4 = (12!)/(8!*4!) = (12*11*10*9)/24 = 495.
Since order does not matter here, it is a combination problem.
Number of ways of selecting 4 sites from given 12 sites is 12C4 = (12!)/(8!*4!) = (12*11*10*9)/24 = 495.
Rahul Lakhani
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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
