Question

A picnic party of 10 persons to go by two vehicles, one matador with a seating capacity of 8 and a maruti car with a seating capacity of 4. In how many ways can the travel arrangements be made?

Answer 1

Hint: Take both vehicles as car 1 and car 2. The maximum capacity is 8 and 4. Consider different cases where 10 people can be arranged like $\left( 8,2 \right),\left( 7,3 \right),\left( 6,4 \right)$ and find the total number of ways using combinations. The formula for combination is given as
\[^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]

Complete step-by-step answer:
It is said that the total number of people in the party is 10. Now these 10 persons are set to travel in two cars. The seating capacity of car 1 = 8. The seating capacity of car 2 = 4. Now let us check into different cases.
i) If 8 people go to by car 1 then 2 people can go by car 2.
ii) If 7 people go by car 1, then 3 people can go by car 2.
iii) If 6 people go by car 1, then 4 people can go by car 2.
The maximum capacity of car 2 is 4, thus we have only 3 cases available. We can write it as,
Car 1(8)
8
7
6
Car 2(4)
2
3
4
Thus total number of ways is
$^{8}{{C}_{8}}{{\times }^{4}}{{C}_{2}}{{+}^{8}}{{C}_{7}}{{\times }^{4}}{{C}_{3}}{{+}^{8}}{{C}_{6}}{{\times }^{4}}{{C}_{4}}$
We know that the formula for combination can be written as $^{n}{{C}_{r}}$ . Here order is not important.
\[\begin{align}
  & ^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} \\
 & {{\therefore }^{8}}{{C}_{8}}=\dfrac{8!}{\left( 8-8 \right)!8!}=1 \\
\end{align}\]
Similarly
$\begin{align}
  & ^{4}{{C}_{4}}=1 \\
 & ^{4}{{C}_{2}}=\dfrac{4!}{\left( 4-2 \right)!2!}=\dfrac{4\times 3\times 2!}{2!\times 2!}=6 \\
 & ^{8}{{C}_{7}}=\dfrac{8!}{\left( 8-7 \right)!7!}=\dfrac{8\times 7!}{7!}=8 \\
 & ^{4}{{C}_{3}}=\dfrac{4!}{\left( 4-3 \right)!3!}=\dfrac{4\times 3!}{3!}=4 \\
 & ^{8}{{C}_{6}}=\dfrac{8!}{\left( 8-6 \right)!6!}=\dfrac{8\times 4\times 6!}{2\times 6!}=28 \\
\end{align}$
Now let us subtract and add the required terms. Total number of ways:
$\begin{align}
  & ^{8}{{C}_{8}}{{\times }^{4}}{{C}_{2}}{{+}^{8}}{{C}_{7}}{{\times }^{4}}{{C}_{3}}{{+}^{8}}{{C}_{6}}{{\times }^{4}}{{C}_{4}} \\
 & =\left( 1\times 6 \right)+\left( 8\times 4 \right)+\left( 28\times 1 \right)=66 \\
\end{align}$
Hence there are a total of 66 ways in which the travel arrangements can be made.

Note: Here the travel arrangements can be made without order which means that the people can be seated wherever they like. Thus we took combination here and not permutation which is more ordered.