Combination!

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Ahmed MS: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Wed Apr 06, 2011 8:35 pm; Followed by:1 members

Combination!

by Ahmed MS » Sun Oct 23, 2011 5:09 am

Q: If 3 of 7 standby passengers are selected for a flight, how many different combinations of standby passengers can be selected?

My answer is 210 (7! / 4!), however the book says 35.

Can anyone help me, please!

Cheers!

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Source: Beat The GMAT — Problem Solving | Sun Oct 23, 2011 5:09 am

user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pm; Thanked: 62 times; Followed by:6 members

by user123321 » Sun Oct 23, 2011 5:13 am

Ahmed MS wrote:Q: If 3 of 7 standby passengers are selected for a flight, how many different combinations of standby passengers can be selected?

My answer is 210 (7! / 4!), however the book says 35.

Can anyone help me, please!

Cheers!

it is asking for combinations not permutations. just selecting any 3 members will do the job. not required to arrange the selected 3 members.

user123321

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Sun Oct 23, 2011 5:43 am

Ahmed MS wrote:Q: If 3 of 7 standby passengers are selected for a flight, how many different combinations of standby passengers can be selected?

My answer is 210 (7! / 4!), however the book says 35.

Can anyone help me, please!

Cheers!

The answer requires us to use combinations (7C3), but your calculations are for permutations.

7C3 = 7!/(3!)(4!)= 35

Or ... to learn a quick way to calculate combinations in your head, you can watch video #17 at: https://www.gmatprepnow.com/module/gmat-counting (it's free)

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com