Chapter 9, Problem 9RE

Solution

To calculate: The number of different license plates that can be formed if a license plate begin with two letters followed by four digits or begin with three digits followed by three letters. Assume that no letters or digits are repeated.

The resultant license plates are $14,508,000$ .

Given information:

Every license plate begins with two letters, followed by four letters or begins with three digits followed by three letters.

Formula used: The number of permutations of n objects taken r at a time denoted by $P_{n_r}$ and given by $P_{n_r}=\frac{n!}{\left(n-r\right)!}$ .

Calculation:

There are 26 letters and 10 digits from 0 to 9.

There are 2 cases, First case says “The first two places are filled with 2 letters and remaining places are filled with 4 digits.”

The first two places can be filled by $P_{{26}_2}$ ways and remaining places can be filled by $P_{{10}_4}$ ways.

The expression can be written as: $P_{{26}_2}\cdot P_{{10}_4}$

Rewrite the expression using the formula $P_{n_r}=\frac{n!}{\left(n-r\right)!}$ ,

  $\begin{array}{c}{P}_{{26}_2}\cdot P_{{10}_4}=\frac{26!}{\left(26-2\right)!}\cdot \frac{10!}{\left(10-4\right)!}\\ =\frac{26!}{24!}\cdot \frac{10!}{6!}\end{array}$

Now expand the factorial,

  $\begin{array}{c}\frac{26\cdot 25\cdot 24!}{24!}\cdot \frac{10\cdot 9\cdot 8\cdot 7\cdot 6!}{6!}=26\cdot 25\cdot 10\cdot 9\cdot 8\cdot 7\\ =3276000\end{array}$

Therefore, the number of license plates in which first two places are filled with 2 letters and remaining places are filled with 4 digits is $3276000$ .

Now consider the second case, “The first three places are filled with 3 digits and remaining places are filled with 3 letters.”

The first three places can be filled by $P_{{10}_3}$ ways and remaining places can be filled by $P_{{26}_3}$ ways.

The expression can be written as: $P_{{}_{{10}_3}}\cdot P_{{26}_3}$

Rewrite the expression using the formula $P_{n_r}=\frac{n!}{\left(n-r\right)!}$ ,

  $\begin{array}{c}{P}_{{10}_3}\cdot P_{{{}_{26}}_{{}_3}}=\frac{10!}{\left(10-3\right)!}\cdot \frac{26!}{\left(26-3\right)!}\\ =\frac{10!}{7!}\cdot \frac{26!}{23!}\end{array}$

Now expand the factorial,

  $\begin{array}{c}\frac{10\cdot 9\cdot 8\cdot 7!}{7!}\cdot \frac{26\cdot 25\cdot 24\cdot 23!}{23!}=720\cdot 15600\\ =11,232,000\end{array}$

Therefore, the number of license plates on second case is $11,232,000$ .

The total number of possible license plates from both cases will become:

  $3,276,000+11,232,000=14,508,000$

Therefore, the number of possible license plates is 14,508,000.