Question

A vehicle registration number consists of 2 letters of English alphabet followed by 4 digits, where the first digit is not zero. Then, the total number of vehicles with distinct registration numbers is
A $ {26^2} \times {10^4} $
B $ {}^{26}{P_2} \times {}^{10}{P_4} $
C $ {}^{26}{P_2} \times 9 \times {}^{10}{P_3} $
D $ {26^2} \times 9 \times {10^3} $

Answer 1

Hint: In order to find the total number of arrangements for registration number, first find the number of arrangements of 2 letter by considering all the alphabets of English can be placed in both the position and then find the number of arrangements for 4 digits by considering 9 ways for first digit and 10 ways for remaining 3 digits. Multiply both the results to obtain the required result.

Complete step-by-step answer:
We are given that a vehicle registration plate consists of 2 letter of English followed by 4 digits, where first digit is not zero
Vehicle registration number= $ \_\,\_\,\_\,\_\,\_\,\_ $
In English alphabet, there are 26 number of letters, so we can say that to fill the first two blanks of registration number we have 26 number of ways for each blank
So the total number of arrangements for 2 letter of English alphabet we have is equal to $ 26 \times 26 = {26^2} $
Now for the 4 digits , we have 10 number of digits ranging from 0,1,2,….9 to put in every digits place .Since we have given a constraint that the first digit is non zero so we have only 9 options available for first digit and 10 for every remaining 3 digits.
So, the total number of arrangements for 4 digits are:
 $ 9 \times 10 \times 10 \times 10 = 9 \times {10^3} $
Now the total number of distinct vehicle registration number is equal to \[
   = \left( {total{\text{ }}no{\text{ }}of{\text{ }}arrangements{\text{ }}of{\text{ }}2{\text{ }}letter} \right) \times \left( {total\,no\,of\,arrangements\,of\,4\,digits} \right) \\
   = {26^2} \times 9 \times {10^3} \;
 \]
Therefore, the correct option is (D) i.e. \[{26^2} \times 9 \times {10^3}\]
So, the correct answer is “Option D”.

Note: 1.Factorial: The continued product of first n natural numbers is called the “n factorial “ and denoted by $ n! $ .
2.Permutation: Each of the arrangements which can be made by taking some or all of a number of things is called as permutations.
If n and r are positive integers such that $ 1 \leqslant r \leqslant n $ , then the number of all permutations of n distinct or different things, taken r at one time is denoted by the symbol $ p(n,r)\,or{\,^n}{P_r} $ .
 $ p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}} $
3.Combinations: Each of the different selections made by taking some or all of a number of objects irrespective of their arrangement is called a combination.
The combinations number of n objects, taken r at one time is generally denoted by
 $ C(n,r)\,or{\,^n}{C_r} $
Thus, $ C(n,r)\,or{\,^n}{C_r} $ = Number of ways of selecting r objects from n objects.
 $ C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $