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Re: Need help on Quants:Combination [#permalink]
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nishadparkhi wrote:
Can anybody explain me the difference between below two combination problem?

1. If seven people board an airport shuttle with only three available seats, how
many different seating arrangements are possible? (Assume that three of
the seven will actually take the seats.)


2.If three of seven standby passengers are selected for a flight, how many
different combinations of standby passengers can be selected?

Thanks in advance.


Hi nishadparkhi, the posters above hit the nail on the head. The only things to verify in combinatorics are whether the order matters and whether the elements can repeat. Once you know those two things, you can either plug the arguments into the formula (permutation if order matters, combination if not) or use logic to sniff out the correct answer.

In the two problems above, the first one actually differentiates between whether Albert, Barry, Charles, David, Edward, Frank or Gerry sit in first class or not. Everyone wants the best seat and no one wants to sit next to the teething infant for 9 hours. In the second case, they just want to get home to their families on the red eye flight, no one is arguing which seat to take, so Albert, Barry and Charles is the same as Charles, Barry and Albert. It's also pretty clear that Albert can't be on the flight three times (unless he's coming from that cloning island in "The Island")... so repetition is not allowed.

Also it's important to keep in mind that when order matters there will be more options than when order doesn't matter. Much the same way A, B and C can be arranged in 6 different orderings, but in the end it's always the same three people on the flight.

n!/(n-k)! for the first one: 7!/4! so that's just 7*6*5 = 210 possibilities. (order matters)

n!/k!(n-k!) for the second one: 7!/4!*3! = 7*6*5/3*2 = 35 possibilities (6 times less, order doesn't matter)

Hope this helps!
-Ron
GMAT Club Bot
Re: Need help on Quants:Combination [#permalink]

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