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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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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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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
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by Ahmed MS » Sun Oct 23, 2011 6:00 am
Sorry, my mistake. Thanks anyway.
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