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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