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Bunuel
If 7 people board an airport shuttle with only 3 available seats, how many different seating arrangements are possible? (Assume that 3 of the 7 will actually take the seats.)

A. 35
B. 180
C. 210
D. 840
E. 5040

7 for first seat*6 for 2nd seat *5 for 3rd seat = 210 IMO C
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Bunuel
If 7 people board an airport shuttle with only 3 available seats, how many different seating arrangements are possible? (Assume that 3 of the 7 will actually take the seats.)

A. 35
B. 180
C. 210
D. 840
E. 5040

total arragements possible ; 7*6*5 ; 210
IMO C
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Bunuel
If 7 people board an airport shuttle with only 3 available seats, how many different seating arrangements are possible? (Assume that 3 of the 7 will actually take the seats.)

A. 35
B. 180
C. 210
D. 840
E. 5040

Arrangements—> Permutation = 7p3
= 7*6*5
= 210

IMO Option C

Pls Hit kudos if you like the solution

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Bunuel
If 7 people board an airport shuttle with only 3 available seats, how many different seating arrangements are possible? (Assume that 3 of the 7 will actually take the seats.)

A. 35
B. 180
C. 210
D. 840
E. 5040


The number of seating arrangements is 7P3 = 7! / (7 - 3)! = 7! / 4! = 7 x 6 x 5 = 210.

Answer: C
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Bunuel
If 7 people board an airport shuttle with only 3 available seats, how many different seating arrangements are possible? (Assume that 3 of the 7 will actually take the seats.)

A. 35
B. 180
C. 210
D. 840
E. 5040

The word "arrangement" clues in that order matters, thus this is permutation. n=7, k=3.

\(\frac{n!}{(n-k)!}=\frac{7!}{4!}=7*6*5=210\)
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