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Number plates of cars must contain 3 letters of the alphabet denoting the place and area to which its owner belongs. This is to be followed by a three – digit number. How many different number plates can be formed if
A) repetition of letters and digits is not allowed
B) rendition of letters and digits is allowed.

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Last updated date: 20th Jun 2024
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Answer
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Hint: We will first make 6 places which are to be used as 3 from alphabets and 3 for digits. Now, fill those spaces up by the number of possibilities which can be filled in them according to both options and then just multiply those possibilities to get the answer.

Complete step-by-step solution:
First since we are given that we must put three alphabets in ahead of the three digits to form one number of the number plate.
Therefore, we have __ __ __ __ __ __ these spaces which need to be filled.
We know that in English letters, we have 26 letters with us and 10 numbers (from 0 to 9).
Part (A):
We need to fill the spaces with use of digits and letters without repetition.
We can fill up the first space by 26 letters, then the second space by 25 letters (because repetition is not allowed), then the third space by 24 letters (because 2 of 26 are already used). Now, we are left to fill 3 blank spaces with digits. So, fourth space can be filled with 10 digits, then fifth place by 9 digits and the last place by 8 digits.
\[\therefore \] We have the number of ways given by = $26 \times 25 \times 24 \times 10 \times 9 \times 8$
Simplifying the calculations, we will get:-
\[\therefore \] We have the number of ways given by = 11232000.
Part (B):
We need to fill the spaces with use of digits and letters with repetition.
We can fill up the first space by 26 letters, then the second space by 26 letters (because repetition is allowed), then the third space by 26 letters (because again repetition is allowed). Now, we are left to fill 3 blank spaces with digits. So, fourth space can be filled with 10 digits, then fifth place by 10 digits and the last place by 10 digits.
\[\therefore \] we have the number of ways given by = $26 \times 26 \times 26 \times 10 \times 10 \times 10$
Simplifying the calculations, we will get:-
\[\therefore \] We have the number of ways given by = 17576000.

Note: The students must keep in mind that if they are not given anything about repetition is not allowed or allowed. We generally assume that it is allowed but if we are given a fixed set of numbers or letters, then we must use among them only, so indirectly repetition will also not be allowed.
We are multiplying the number of possibilities of letters and digits with each other, because these all are possibilities corresponding to each other.
Alternate way:-
You may choose 3 letters and 3 digits out of 26 letters and 10 digits respectively by using the formula $^{26}{C_3}$ and $^{10}{C_3}$ (in case of without repetition) and then permute those digits and letters by multiplying by 3! each of them and thus you will get the same answer.