Question

Number plates of cars must contain 3 letters of alphabet denoting the place and area to which its owner belongs. This is to be followed by a 3 digit number. How many different number plates can be formed if
(i)Repetition of letters and digits is not allowed
(ii)Repetition of letters and digits is allowed

Answer 1

Hint: This is a problem related to Permutation and Combination. Here, it is noted that we have to arrange the numbers and alphabets in order; therefore we will apply Permutation fundamentals. Combination fundamentals are used where we have to select the numbers, objects etc., from the group of objects or collection.

Complete step-by-step answer:
From the problem, we have to arrange the alphabets and numbers in a specific order. We also know that there are 26 letters in alphabets and 10 digits in numbers.
Now, there are two cases:
Case (i): In this case, we are not allowed to repeat the letters and digits.
Now, to choose 3 letters from 26 alphabets, the permutation formula will be
${}^{26}{P_3}$ ---------- (1)
This can be expanded using factorial fundamental, which says that if$r$objects are to be chosen from $n$objects without repetition then,
\[{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}\] ……………… (2)
Now, with the help of above explanation, expression (1) can be expanded as below,
${}^{26}{P_3} = \dfrac{{26!}}{{(26 - 3)!}}$
This can further be simplified as
$
   = \dfrac{{26!}}{{(26 - 3)!}} \\
   = \dfrac{{26 \times 25 \times 24 \times 23!}}{{23!}} \\
   = 26 \times 25 \times 24 \\
   = 15600{\text{ }}...................................{\text{(3)}} \\
 $
In a similar way, we will calculate to choose 3 digits from 10 digits. It is to be remembered here that zero is also a digit but when it comes to first place the number will become two digit numbers. To avoid it, we will deduct the 3 digit numbers which are having zero at their first place. This can be done as below using the expression (2):
$
   = {}^{10}{P_3} - {}^9{P_2} \\
   = \dfrac{{10!}}{{(10 - 3)!}} - \dfrac{{9!}}{{(9 - 2)!}} \\
   = \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!}} - \dfrac{{9 \times 8 \times 7!}}{{7!}} \\
 $
This can further be simplified as
$
   = \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!}} - \dfrac{{9 \times 8 \times 7!}}{{7!}} \\
   = 10 \times 9 \times 8 - 9 \times 8 \\
   = 9 \times 8 \times (10 - 1) \\
   = 9 \times 8 \times 9 \\
   = 648{\text{ }}..........................................{\text{(4)}} \\
 $
Therefore total number of plates can be calculated from expression (3) and (4)
$
   = 15600 \times {\text{648}} \\
  {\text{ = 10108800}} \\
 $
Case (ii): In this case, we are allowed to repeat the letters and digits.
Therefore, all three places can be filled with all 26 letters, thus,
Total numbers will be
$
   = 26 \times 26 \times 26 \\
   = 17576 \\
 $ ………………….. (5)
And all three places can be filled with all 10 digits. It is to be remembered here that zero is also a digit but when it comes to first place the number will become two digit numbers. To avoid it, we will deduct the 3 digit numbers which are having zero at their first place. This can be expressed as
$
   = (10 - 1) \times 10 \times 10 \\
   = 9 \times 10 \times 10 \\
   = 900 \\
 $………………….. (6)
Therefore total number of plates can be calculated from expression (5) and (6)
$
   = 17576 \times 900 \\
  {\text{ = 15818400}} \\
 $

Note: Solving problems related to digits, you should remember that if zero comes to the first place a number it will become a number of one digit less than what we are expecting. And remember that where to use permutation and where to use combination fundamentals.