How many 4-letter code can be formed using the first 10 letters of the English alphabet, if (i) no letter is r

Exercise 6.2 · Q3

Q. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if

(i) no letter is repeated

(ii) repetition of letters is allowed

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Concept understanding — Factorial

What is a Factorial?

Imagine you have 3 different books and you want to arrange them on a shelf. How many different ways can you do it?

You could try listing them: Book A, B, C — or A, C, B — or B, A, C — and so on. If you actually do this, you'll find exactly 6 arrangements.

That "6" is no accident. It comes from multiplying: $3 \times 2 \times 1 = 6$ .

This pattern — multiplying a number by every positive integer smaller than it, all the way down to 1 — is called a factorial. We write it with an exclamation mark: $3! = 3 \times 2 \times 1 = 6$ .

Note

The exclamation mark is not for excitement — it's just the notation. $5!$ is read as "five factorial", not "five!" with surprise.

The Precise Definition

For any positive integer $n$ , the factorial $n!$ is defined as:

n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1

So:

$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Notice how fast factorials grow. $5!$ is already 120, and $10!$ is 3,628,800.

The Special Case: $0!$

This often confuses students. What is $0!$ ?

By the definition above, $0!$ would be an empty product — multiplying nothing. That doesn't make sense directly, so we define it by what makes the mathematics consistent.

Consider this pattern:

\frac{5!}{5} = 4! \quad \Rightarrow \quad \frac{120}{5} = 24

\frac{4!}{4} = 3! \quad \Rightarrow \quad \frac{24}{4} = 6

\frac{3!}{3} = 2! \quad \Rightarrow \quad \frac{6}{3} = 2

\frac{2!}{2} = 1! \quad \Rightarrow \quad \frac{2}{2} = 1

\frac{1!}{1} = 0! \quad \Rightarrow \quad \frac{1}{1} = 1

So $0! = 1$ . This isn't a guess — it's forced by the pattern, and it makes all formulas involving factorials work correctly.

Important

$0! = 1$ — memorize this. It is not zero.

Why Do We Need Factorials?

Factorials are the building block for counting arrangements and selections. The two most common uses are:

Permutations — arranging $n$ distinct objects in order: $n!$ ways.

Combinations — choosing $r$ objects from $n$ without caring about order: $\frac{n!}{r!(n-r)!}$ .

You'll see these in probability, statistics, and algebra (binomial theorem). For now, just get comfortable with computing factorials and understanding that $n!$ counts the number of ways to order $n$ distinct items.

A Quick Check

Try these yourself:

$6! = ?$
$\frac{7!}{5!} = ?$
$4! + 3! = ?$

Answers: 720, 42, 30. If you got them, you've understood the core idea.

A 4-letter code from the first 10 letters (A–J) gives $^{10}P_4=5040$ codes without repetition, and $10^4=10000$ codes with repetition allowed.

[!FORMULA] $^{n}P_{r}=\dfrac{n!}{(n-r)!}$ counts arrangements (order matters) of $r$ items chosen from $n$ distinct items without repetition; when repetition IS allowed, each of the $r$ positions independently has $n$ choices, giving $n^{r}$ .

We have $n=10$ letters (first 10 letters of the alphabet) and code length $r=4$ .
(i) No letter repeated: the code is an ordered selection without repetition, so it is $^{10}P_4$ .

$^{10}P_4=\dfrac{10!}{6!}=10\times9\times8\times7$ .

Multiply: $10\times9=90$ , $90\times8=720$ , $720\times7=5040$ .

So $^{10}P_4=5040$ .

(ii) Repetition allowed: each of the 4 positions can independently be any of the 10 letters, so the count is $10\times10\times10\times10=10^4$ .
Compute: $10^4=10000$ .
Self-check: without repetition the count must be less than with repetition, and $5040<10000$ ✓ (also $5040=10\times9\times8\times7<10\times10\times10\times10=10000$ , consistent).

✓Final answer

5040

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Q.How many 4-letter code can be formed using the first 10 letters of the English alphabet, if

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