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

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

00:00

A

B

C

D

E

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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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