How many automobile license plates can be made if each plate contains taw different letters followed by three different digits. Solve the problem if the first digit can't be 0.

Solution:

$\mathrm{I}^{\text {st }}$ position is a letter and hence can be selected from 26 letters.

$$ ={ }^{26} \mathrm{C}_{1} \text { ways } \\ $$

II $^{\text {ad }}$ position is also a letter but different from $I^{\text {st }}$ letter can be selected from 25 letters,

$$ ={ }^{25} \mathrm{C}_{1} \text { ways } \\ $$

Now in digits, It $^{\text {th }}$ position can be selected from 10 digits

$$ ={ }^{10} \mathrm{C}_{1} \text { ways } \\ $$

$$ ={ }^{9} \mathrm{C}_{1} \text { ways } \\ $$

And III$^{\text {ad }}$ position can be selected from 8 digits as it is different from $I^{\text {st }}$ and $\mathrm{I}^{\text {rd }}$ digit,

$$ ={ }^{8} C_{1} \text { ways } \\ $$

Hence, the total number of License plates,

$$ \begin{aligned} &=26 \times 25 \times 10 \times 9 \times 8 \\\\ &=468000 \\ \end{aligned} $$

Now in License plate $I^{\text {st }}$ digit cannot be zero, then $I^{\text {st }}$ position can be selected out of 9 digits, ie. in ${ }^{9} \mathrm{C}_{1}$ ways.

II $^{\text {nd }}$ position can have zero, but one digit is already selected for $1^{\text {st }}$ position hence $2^{\text {nd }}$ digit can be selected in ${ }^{9} \mathrm{C}_{1}$ ways also III$\mathrm{}^{\text {rd }}$ digit can be selected in ${ }^{8} \mathrm{C}_{1}$ ways.

Hence, the total number of auto license plates,

$$ =26 \times 25 \times 9 \times 9 \times 8=421200 \\ $$