Joan is allowed to invite 3 of her friends to join her on a family camping trip. If Joan has 10 friends, in how many way

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Tue Feb 16, 2021 7:42 pm

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Joan is allowed to invite 3 of her friends to join her on a family camping trip. If Joan has 10 friends, in how many ways can she invite 3 of them?

A. 27
B. 120
C. 140
D. 360
E. 720

OA B

Source: Magoosh

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Source: Beat The GMAT — Problem Solving | Tue Feb 16, 2021 7:42 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

Re: Joan is allowed to invite 3 of her friends to join her on a family camping trip. If Joan has 10 friends, in how many

by deloitte247 » Sun Feb 21, 2021 12:13 pm

Joan's Total Friends = 10
Friends to select = 3
Selecting 3 out of 10 friends $$10C_3=\frac{10!}{3!\left(10-3\right)}=\frac{10\cdot9\cdot8\cdot7!}{3!\cdot7!}=10\cdot3\cdot4=120$$
$$Answer\ \ is\ Option\ B$$

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8175; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: Joan is allowed to invite 3 of her friends to join her on a family camping trip. If Joan has 10 friends, in how many

by Scott@TargetTestPrep » Tue Feb 23, 2021 3:44 pm

BTGmoderatorDC wrote: ↑
Tue Feb 16, 2021 7:42 pm
Joan is allowed to invite 3 of her friends to join her on a family camping trip. If Joan has 10 friends, in how many ways can she invite 3 of them?

A. 27
B. 120
C. 140
D. 360
E. 720

OA B

Solution:

The order in which she invites her friends is of no consequence; hence, we will use combinations instead of permutations. The number of ways to invite 10 friends from 3 is:

10C3 = 10! / (3! x 7!) = (10 x 9 x 8) / 3! = 5 x 3 x 8 = 120.

Answer: B

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Founder and CEO
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