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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The three competitors on a race have to be randomly... tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 0 member replies ### Top Member ## The three competitors on a race have to be randomly... The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios content at least one woman? A. 10 B. 15 C. 16 D. 30 E. 46 The OA is E. I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 LUANDATO wrote: The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios content at least one woman? A. 10 B. 15 C. 16 D. 30 E. 46 The OA is E. I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced. One approach: find the number of ways three competitors can be chosen without restriction and then subtract out the number of undesirable outcomes. Total # of ways to select 3 people from a group of 8 with no restrictions: 8C3 = 8*7*6/3! = 56 An undesired outcome, if we want at least one woman, would be to have no women in the group. The number of ways we can select 3 men from a group of 5: 5C3 = 5*4*3/3! = 10 # Total - #undesired = 56 - 10 = 46. The answer is E _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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### GMAT/MBA Expert

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Hi LUANDATO,

We're told to choose three competitors from a group of five men and three women. We're asked for the number of different trios that include AT LEAST one woman. The answer choices to this question are 'spaced out' enough that you can do a little bit of work to eliminate all of the wrong answers.

To start, if the group has just 1 woman, then there will be 2 men. The number of groups of 2 men is 5c2 = 5!/(2!)(3!) = 10 different pairs of men.

With 3 women to choose from, there would be 3(10) = 30 groups with just 1 woman. There would then be additional groups with 2 women or all 3 women, so the total number of groups MUST be greater than 30. There's only one answer that's possible...

GMAT assassins aren't born, they're made,
Rich

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