The three competitors on a race have to be randomly...

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

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

The three competitors on a race have to be randomly...

by BTGmoderatorLU » Fri Nov 17, 2017 11:19 am

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.

Quote

Source: Beat The GMAT — Problem Solving | Fri Nov 17, 2017 11:19 am

DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 am; Location: Boston, MA; Thanked: 1153 times; Followed by:128 members; GMAT Score:770

by DavidG@VeritasPrep » Fri Nov 17, 2017 11:25 am

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

Quote

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Fri Nov 17, 2017 11:41 am

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...

Final Answer: E

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

Contact Rich at [email protected]

Quote

GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8087; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sat Oct 26, 2019 7:49 am

BTGmoderatorLU 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.

We can use the formula:

Total number of ways to select the group - number of ways with no women = number of ways with at least one woman.

Total number of ways to select 3 people from a group of 8 people:

8C3 = 8!/[3!(8-3)!] = 8!/3!5! = (8 x 7 x 6)/3! = 56

Number of ways with no women:

5C3 = 5!/[3!(5-3)!] = (5 x 4 x 3)/3! = (5 x 4 x 3)/(3 x 2 x 1) = 10

Thus, the number of ways with at least one woman is 56 - 10 = 46.

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]