Permutation-Combination
Aptitude

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Q.

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.

9

 B.

26

 C.

126

 D.

392

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Solution:
Option(C) is correct

There are $8$ students and the maximum capacity of the cars together is $9$.

We may divide the $8$ students as follows

Case I: $5$ students in the first car and $3$ in the second 

Or

Case II: $4$ students in the first car and $4$ in the second

Hence, in Case I: $8$ students are divided into groups of $5$ and $3$ in${^8C_3}$ ways.

Similarly, in Case II: $8$ students are divided into two groups of $4$ and $4$ in ${^8C_4}$ ways.

Therefore, the total number of ways in which $8$ students can travel is:

$={^8C_3} + {^8C_4}$

$= 56 + 70$

$= \textbf{126}$


(6) Comment(s)


Akshay
 ()

Saying that 5 students are seated in the first car and 3 in the second, why the condition that the 5 students sitting in the first car can be seated among themselves in 5! ways is not considered??

Similarly why is that the 3 students sitting in the second car can be arranged in 4C3 ways is not considered?? The same logic applies for case 2 also.



May
 ()

Why aren't the fact that those students could sit anywhere is not discussed?


Preeti
 ()

Students can sit anywhere and terms, ${^8C_3}+{^8C_4}=126$ reflect this. Why do you feel its's not discussed?

Can you elaborate more?


Suyog
 ()

could you please explain me how ${^8C_3}$ and ${^8C_4}$ came from case I and II respectively?


Deepti
 ()

Case I says $5$ students in the first car and $3$ in the second. So it becomes choosing $3$ students from $8$ students or $^8C_3$ choices.

Similarly Case II says $4$ students in the first car and $4$ in the second. So it becomes choosing $4$ students from $8$ students or $^8C_4$ choices.

Thus total choices,

$={^8C_3}+{^8C_4}=126$

Payal
 ()

you can do it in this way

8C5*3C3+8C4*4C4 =126