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John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different

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by BTGmoderatorDC » Mon Jun 27, 2022 12:04 am

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John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A) 3
B) 8
C) 10
D) 15
E) 20


OA C

Source: GMAT Prep
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Re: John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many differ

by swerve » Mon Jun 27, 2022 9:39 am
BTGmoderatorDC wrote: ↑
Mon Jun 27, 2022 12:04 am
 John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A) 3
B) 8
C) 10
D) 15
E) 20


OA C

Source: GMAT Prep
Let's try as follows,

There are only \(3\) places\(: + + + \)
But \(5\) friends that signifies \(2\) friends without places\(: x \, x\)

Hence we can represent\(: + + + \, x \, x \)

That is \(3!\) and \(2!\) our denominator

Total \(5\) friends or \(5!\) is a numerator

\(\dfrac{5!}{3!\cdot 2!} = 10\)

Hope it helps
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Re: John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many differ

by Scott@TargetTestPrep » Fri Jan 27, 2023 8:22 am
BTGmoderatorDC wrote: ↑
Mon Jun 27, 2022 12:04 am
 John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A) 3
B) 8
C) 10
D) 15
E) 20


OA C

Source: GMAT Prep
The number of different combinations of 3 passengers can be formed from the 5 people is 5C3 = (5 x 4 x 3) / (3 x 2) = 60/6 = 10.

Answer: C

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