How many automobile license plates can be made if each plate contains two different letters followed by three

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Q. How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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The problem asks for the number of license plates with two different letters followed by three different digits. This is a direct application of the multiplication principle with permutations without repetition: $26 \times 25 \times 10 \times 9 \times 8 = 468,\!000$ .

The key idea here is selection without replacement. When a plate has "two different letters," it means once you pick the first letter, you cannot use it again for the second spot. The same logic applies to the three digits — each must be distinct from the others.

This is not a combination problem, because the order matters: "AB" is a different plate from "BA". So we are counting permutations of a subset of items, not combinations.

Let’s break it down step by step.

Choose the first letter.

There are 26 letters in the English alphabet. No restrictions yet, so we have 26 choices.
Choose the second letter, which must be different from the first.

After picking the first letter, only 25 letters remain. So there are 25 choices here.
Choose the first digit.

Digits are 0 through 9, giving 10 possibilities. No restrictions yet.
Choose the second digit, which must be different from the first.

One digit is already used, so 9 choices remain.
Choose the third digit, which must be different from both previous digits.

Two digits are already taken, so 8 choices remain.

Now, by the multiplication principle (if one event can happen in $m$ ways and a subsequent independent event in $n$ ways, the total number of outcomes is $m \times n$ ), we multiply all these choices together:

26 \times 25 \times 10 \times 9 \times 8

Let’s compute stepwise:

$26 \times 25 = 650$
$10 \times 9 = 90$
$90 \times 8 = 720$ …

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Q.How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

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