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# Need help on Quants:Combination

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Intern
Joined: 04 Mar 2013
Posts: 2

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Need help on Quants:Combination [#permalink]

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14 Mar 2013, 01:35
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.

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VP
Joined: 02 Jul 2012
Posts: 1216

Kudos [?]: 1691 [3], given: 116

Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Need help on Quants:Combination [#permalink]

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14 Mar 2013, 04:40
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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?

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.

For ease of illustration I'll make 7 into 5 and 3 into 2.

So the questions would now be

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

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

The answer to the first question would be $$5C_2$$ = $$\frac{5*4}{1*2}$$ = 10

The answer to the second question would be $$5P_3$$ = 5*4 = 20

Illustration :

Let the five people be A,B,C,D,E

The different groups of two that can be formed from this five are
AB,BA,
AC, CA,
AD, DA,
AE, EA,
BC, CB,
BD, DB,
BE, EB,
CD, DC,
CE, EC,
DE, ED

The first question asks for "seating arrangements". If there are two chairs side by side, then AB would be a different seating arrangement from BA. Hence for this question AB and BA are considered as two different groups.

The second question just asks us to select two people from five. We do not care about what order they stand in. So in this case AB and BA are both the same group.

Hope it's clear.
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Re: Need help on Quants:Combination [#permalink]

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14 Mar 2013, 04:45
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The first one asks you to calculate the permutations (different arrangements). e.g. if there were 2 seats and 2 people A and B, then there will be 2 methods to seat them. AB and BA (A takes the first seat or B takes the first seat).

The second one asks you to calculate the combinations (different selections and not arrangements). So, for 2 people and 2 seats, number of ways will be 1 - AB/BA (doesn't matter who sits on the first seat. We just want to know in how many ways you can 'select' the 2 people to sit)

For more explanations on P&C theory you may download the GMAT club math book.

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Re: Need help on Quants:Combination [#permalink]

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14 Mar 2013, 07:40
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Expert's post
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
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Re: Need help on Quants:Combination   [#permalink] 14 Mar 2013, 07:40
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