BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Permutation and combinations

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 98
Joined: Mon Nov 23, 2009 2:30 pm
Thanked: 26 times
Followed by:1 members

by ace_gre » Fri Jan 15, 2010 6:03 pm
Hi, Please post the answer choices, as this makes it easier to solve..

Here is my approach.

No. of ways for selecting (>= 2 JKQ) = Total ways of selecting 3 cards - # of ways of selecting (<2 JKQ) = Total - (zero JKQ) -P( one JKQ)

Total ways of selecting 3 cards = 52C3
No. of JKQ = 12
Non JKQ cards = 52-12 = 40
Ways of selecting 0 JKQ = 40C3
Ways of selecting 1 JKQ = 12C1 * 40C2

Ways to select >= two JKQ ==> 52C3 - 40C3- 12* 40C2

Is there anything along these lines in the choices?
Top
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Fri Jan 15, 2010 4:18 pm

by hemanthquartz1 » Fri Jan 15, 2010 7:18 pm
You are right. Answer is 2860 :)

Thanks
Top
Post new topic Post Reply
• Page 1 of 1