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

Hint: 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 \[^8{C_5}{ = ^8}{C_3} = \dfrac{{8!}}{{3!5!}}\]