A team of 8 students goes on an excursion, in two cars, of which one

Question

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

A. 120
B. 126 
C. 146 
D. 156
E. 166

ScottTargetTestPrep · Accepted Answer

We can call the 5-seat car Car A and the 4-seat car Car B. Therefore, we have two cases to consider:

Case 1: Car A seats 5 and Car B seats 3

Case 2: Car A seats 4 and Car B seats 4

Case 1:

If Car A seats 5 and Car B seats 3, then we have 8C5 x 3C3 = (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2) x 1 = 8 x 7 = 56 ways for the 8 students to travel in these two cars.

Case 2:

If Car A seats 4 and Car B seats 4, then we have 8C4 x 4C4 = (8 x 7 x 6 x 5)/(4 x 3 x 2) x 1 = 2 x 7 x 5 = 70 ways for the 8 students to travel in these two cars.

So we have a total of 56 + 70 = 126 ways for the 8 students to travel in these two cars.

Answer: B.

chetan2u · Answer

Hi, Archit3110, the way would be 
We have 8 students and the question finally boils down to ways to distribute 8 as 5 and 3 in one grouping and as 4 and 4 in other grouping..
so 8C5 and the remaining three will form a group of 3, and 8C4 and other 4 will automatically make the group of remaing 4..
so 8C5+8C4 =  \frac{8*7*6}{3*2}+\frac{8*7*6*5}{4*3*2} =56+70=126

Archit3110 · Answer

total students = 8
total available seats are 9
so 
8c5+8c4=56+70=126
IMOB

gvij2017 · Answer

Isn't it a permutation problem?
How many ways 8 people can travel in 2 cars- 5 seats in one car and 4 in another.

tati912 · Answer

There are 8 students but the available total seats are 9. There will always be one vacant seat in one of the cars. So we're really finding how many ways we can group the 9.

9! / 5! 4! = 126

Posted from my mobile device

luisdicampo · Answer

Deconstructing the Question  
There are 8 students and two distinct cars:
Car A can seat up to 5, Car B can seat up to 4.
We count assignments of students to cars (not specific seats).
Car A does not have to be full, so two valid split cases exist: 5+3 and 4+4.

Step-by-step  
Case 1: Car A has 5 and Car B has 3
 \binom{8}{5}=56

Case 2: Car A has 4 and Car B has 4
 \binom{8}{4}=70

Total:
 56+70=126

Answer: B

A team of 8 students goes on an excursion, in two cars, of which one

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A team of 8 students goes on an excursion, in two cars, of which one

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