Question

How many number plates can be made if the number plates have two letters of the English alphabet (a-z) followed by two digits (0-9) if the repetition of digits or alphabets is not allowed?
(A) $58500$
(B) $54000$
(C) $35500$
(D) $50000$

Answer 1

To solve this sum, we use the concepts of permutations and combinations. Selecting things from given things is known as combination and arranging the given things is known as permuting. Also, if there are $n$ things which have to be arranged in $n$ places, then number of ways of doing them is $n!$
And we also use a formula which is ${}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}$ .

Complete step-by-step solution:
In the question, it is given that a number plate has two letters and two numbers. So, the number plate will be in this way: -
$XY{\text{ 15}}$
Suppose that, if you have $m$ shirts and $n$ pants, then the number of different ways of wearing a pair is equal to $m \times n$ . We use this concept to solve this problem.
First, let us calculate number of two-digit alphabet that can be made from 26 alphabet (A-Z)
If we have two things, say A and B, then these can be arranged in two ways which are AB or BA.
So, any two things form a pair, and one pair can be arranged in two ways. ----(1)
Number of ways of selecting $n$ things from $m$ things is equal to ${}^m{C_n}$ . -----(2)
Now, we have 26 alphabets, and we have to select two from them.
So, the number of combinations that we get is equal to ${}^{26}{C_2}$ .
One combination can be arranged in two ways (from (1)), so, ${}^{26}{C_2}$ combinations can be arranged in ${}^{26}{C_2} \times 2$ ways. -----(3)
Now, let us find number of two-digit numbers that can be formed from digits 0-9
Here also, we have 10 digits and we have to take two out of them. So, the number of combinations will be ${}^{10}{C_2}$ . And, these combinations can be arranged in ${}^{10}{C_2} \times 2$ ways. ----(from (1))
Now we have ${}^{26}{C_2} \times 2$ ways of two alphabets and ${}^{10}{C_2} \times 2$ ways of two numbers.
So, total number of combinations will be equal to $({}^{26}{C_2} \times 2) \times ({}^{10}{C_2} \times 2)$
And we know that, ${}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}$ .
So,
$\Rightarrow ({}^{26}{C_2} \times 2) \times ({}^{10}{C_2} \times 2) = \dfrac{{26!}}{{2!\left( {26 - 2} \right)!}} \times 2 \times \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} \times 2$
$= \dfrac{{26!}}{{2!(24)!}} \times \dfrac{{10!}}{{2!\left( 8 \right)!}} \times 4$
And we know that $n! = n(n - 1)! = n(n - 1)(n - 2)!$ and so on.
$= \dfrac{{26(25)(24)!}}{{2!\left( {24} \right)!}} \times \dfrac{{10(9)(8)!}}{{2!\left( 8 \right)!}} \times 4$
$= \dfrac{{26(25)}}{{2!}} \times \dfrac{{10(9)}}{{2!}} \times 4$
And also, 2!$$$$ = 2 \times 1 = 2
$= \dfrac{{650}}{2} \times \dfrac{{90}}{2} \times 4$
$= 650 \times 90$
$= 58500$
So, number plates can be made if the number plates have two letters of the English alphabet (a-z) followed by two digits (0-9) is equal to $58500$ . So, option (A) is the correct option.

Note: If there are $m$ things out of which $n$ things have to be arranged then number of possible ways is equal to ${}^m{P_n} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ . This is known as permutation. Also remember that, in a permutation ${}^m{P_n}$ or in a combination ${}^m{C_n}$ , $n$ is always less than $m$ . If it is greater, then the value cannot be evaluated. And also remember the relation, ${}^m{C_n} \times r! = {}^m{P_n}$ .