Question

You need to have a password with

4

letters followed by

4

even digits between

0

and

9

, inclusive. If the characters and digits cannot be used more than once, how many choices do you have for your password?

Answer 1

Concepts and reason

The concept of permutations is used to solve this problem.

Permutations are one of the methods to calculate the total outcomes of an event if the order of the outcomes is considered.

Fundamentals

The arrangement of $n$ objects can be calculated by using the formula:

${\rm{Number of possible arrangement}} = n!$

There are $n$ objects, and if $r$ objects are selected at a time when the repetition is not allowed, then the number of possible selections is,

$\begin{array}{c}\\{\rm{Number of possible selection}} = {}^n{P_r}\\\\ = \frac{{n!}}{{\left( {n - r} \right)!}}\\\end{array}$

The password should have 4 letters, which must be followed by 4 even digits between 0 and 9, both inclusive. The letters or the digits should not be repeated.

As there are 26 alphabets, there are ${}^{26}{P_1}$ ways of selecting the first letter. As the letters are not repeated, there are ${}^{25}{P_1}$ ways of selecting the second letter. In a similar way, the third and the fourth letter can be selected in ${}^{24}{P_1}$ and ${}^{23}{P_1}$ ways.

There are 5 even digits $\left\{ {0,2,4,6,8} \right\}$ between 0 and 9, both inclusive, and 4 digits have to be selected. So, there are ${}^5{P_1}$ ways of selecting the first digit. As the digits are also not repeated, there are ${}^4{P_1}$ ways of selecting the second digit. In a similar way, the third and the fourth digit can be selected in ${}^3{P_1}$ and ${}^2{P_1}$ ways.

The number of choices for the password of 4 letters and 4 digits is calculated as:

$\begin{array}{c}\\{\rm{Number of choices}} = \left( {{}^{26}{P_1} \times {}^{25}{P_1} \times {}^{24}{P_1} \times {}^{23}{P_1} \times {}^5{P_1} \times {}^4{P_1} \times {}^3{P_1} \times {}^2{P_1}} \right)\\\\ = \left( \begin{array}{l}\\\frac{{26!}}{{\left( {26 - 1} \right)!}} \times \frac{{25!}}{{\left( {25 - 1} \right)!}} \times \frac{{24!}}{{\left( {24 - 1} \right)!}} \times \frac{{23!}}{{\left( {23 - 1} \right)!}}\\\\ \times \frac{{5!}}{{\left( {5 - 1} \right)!}} \times \frac{{4!}}{{\left( {4 - 1} \right)!}} \times \frac{{3!}}{{\left( {3 - 1} \right)!}} \times \frac{{2!}}{{\left( {2 - 1} \right)!}}\\\end{array} \right)\\\\ = 26 \times 25 \times 24 \times 23 \times 5 \times 4 \times 3 \times 2\\\\ = 43056000{\rm{ ways}}\\\end{array}$

Ans:

The number of choices for the password is 43,056,000.