Number of ways in which 8 person can be seated in three different taxies each having 3 seats for passenger and duly numbered. internal arrangement matters.
My approach:
Method (1):- Since seats are numbered, we can consider all seats to be different. I want to seat eight persons on $9$ seats, which can be done by $C(9,8) \cdot 8! = 9!$.
Method (2):- First, I'll make $3$ unnumbered group of sizes $3,3,2$, which can be done in $\frac {8!}{3! \cdot 3! \cdot 2! \cdot 2!}$ ways. Now I will distribute these group to $3$ taxis, which can be done in $3!$ ways. And internal arrangement of seats in the taxis can be done in $3!$, $3!$, $2!$. So final answer will be $$\frac {8!}{3! \cdot 3! \cdot 2! \cdot 2!} \cdot {3! \cdot 3! \cdot 3! \cdot 2!}$$ That is equal to $3 \cdot 8!$.
What is going wrong in second method?
I want to know mistake in second problem.
Best Answer
There are three seats available in the taxi which holds only two people. The two people in that taxi can be seated in $3! = 3 \cdot 2$ ways since there are three ways to seat the younger person and two ways to seat the older person. Replacing the factor of $2!$ by $3!$ in your second method yields $$\frac{8!}{3!3!2!2!} \cdot 3! \cdot 3! \cdot 3! \cdot 3! = 8! \cdot 3 \cdot 3 = 9!$$ in agreement with the correct answer you obtained using your first (simpler) method.