Question

A group of $8$ students go for a picnic in two cars, of which one can seat $5$ and the other only $4$. In how many ways can they travel?

Answer 1

Let us consider, total number of objects is $n$. Among $n$ objects, $r$ objects can be arranged in $^n{C_r}$ ways.
That is we are going to solve the given problem using a combination of objects.

Complete step-by-step answer:
It is given that a group of $8$ students go for a picnic in two cars of which one can seat $5$ and the other only $4$. We have to find the ways they can travel in both cars.
There are $8$ students and the maximum capacity of the cars together is $5 + 4 = 9$
So, we can divide $8$ students in two ways.
Case 1: $5$ students are in one car and $3$ students in the other.
$8$ students are divided into two groups of $5$ students and $3$ students can be arranged in $8{C_3}$ ways.
Case 2: $4$ students are in one car and $4$ students in the other.
$8$ students are divided into two groups of $4$ students and $4$ students can be arranged in $8{C_4}$ ways.
Therefore, the total number of ways in which $8$ students can be travelled is:
$= 8{C_3} + 8{C_4}$
Combination formula is $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Let us solve the above equation using combination formula we get,
$= \dfrac{{8!}}{{3!5!}} + \dfrac{{8!}}{{4!4!}}$
By simplifying the factorials in the above equation we get,
$= \dfrac{{8 \times 7 \times 6}}{{3 \times 2}} + \dfrac{{8 \times 7 \times 6 \times 4}}{{4 \times 3 \times 2}}$
Let us cancel the common terms in numerator and denominator and multiplying the remaining terms, we get,
$= 56 + 70 = 126$
Hence, they can travel in $126$ ways.

Note: For first case, if we consider, $8$ students are divided into two groups of $5$ students and $3$ students can be arranged in $^8{C_5}$ ways, then also the number of ways that they can travel is,
${ = ^8}{C_5}{ + ^8}{C_4} = 126$
Here we can say $^8{C_5}{ = ^8}{C_3}$ using combination formula $^8{C_5}{ = ^8}{C_3} = \dfrac{{8!}}{{3!5!}}$