Question

Number of ways in which $8$ persons can be seated in $3$ different taxis, each having $3$ seats for passengers and seats are numbered if the internal arrangements of persons inside the taxi does not matter?

Answer 1

In the given question we are required to find out the number of ways in which a certain number of people can be seated in a specific number of taxis. Each taxi has a particular number of seats. The internal arrangements of persons inside the taxi does not matter. So, we just have to find ways of assigning people to one of the taxis. We use the fundamental theorem of counting to solve the problem.

Complete step by step answer:
We are required to find the number of ways in which $8$ persons can be seated in $3$ different taxis, each having $3$ seats for passengers.So, there are a total of $9$ seats.
Number of passengers $= 8$
Also, the internal arrangements of persons inside the taxi does not matter. So, we just have to assign a taxi to each of the passengers.Now, we have three different numbered taxis, nine seats and eight passengers. So, one seat in one of the taxis will be left vacant. We also know that the formula for selecting $r$ things out of $n$ different things is $^n{C_r}$. The expansion for the combination formula is $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$.

So, the number of options for choosing a taxi with only two people in it is $^3{C_1} = 3$.Now, we have to select two passengers for the taxi with one vacant seat. So, the number of options for choosing two people out of eight is $^8{C_2} = \dfrac{{8!}}{{6! \times 2!}} = \dfrac{{8 \times 7}}{2} = 28$.
Now, the remaining two taxis will carry three passengers each.
So, the number of ways of selecting three people out of the remaining six is,
$^6{C_2} = \dfrac{{6!}}{{2! \times 4!}} \\\ \Rightarrow ^6{C_2}= \dfrac{{6 \times 5}}{2} \\\ \Rightarrow ^6{C_2} = 15$

Now, the remaining three people will sit in the third taxi. So, the number of ways of selecting three out of three people is $^3{C_3} = 1$.
Hence, we get the number of ways as,
$\Rightarrow 3 \times 28 \times 15 \times 1$
Carrying out the calculations, we get,
$\Rightarrow 84 \times 15$
$\Rightarrow 1260$
Therefore, the number of ways in which $8$ persons can be seated in $3$ different taxis, each having $3$ seats for passengers and seats are numbered if the internal arrangements of persons inside the taxi does not matter is $1260$.

Note: The question revolves around the concepts of Permutations and Combinations. One should know about the principle rule of counting or the multiplication rule. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer. We must know how to calculate the value of the combination formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$.