Is 3! + 4! = 7!?

Exercise 6.2 · Q2

3! + 4! = 7!

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Concept understanding — Factorial Arithmetic

Factorial Arithmetic — From Intuition to Precision

Imagine you have 3 different books you want to arrange on a shelf. How many different ways can you line them up? You could try listing them: Book A, B, C — or A, C, B — or B, A, C — and so on. If you actually count, you'll find 6 arrangements.

Where does that 6 come from? For the first position, you have 3 choices. Once you pick one, you have 2 choices left for the second position. Then only 1 choice remains for the last spot. So the total is $3 \times 2 \times 1 = 6$ .

That product — multiplying a whole number by every positive integer smaller than it, all the way down to 1 — is called a factorial. It's one of the most useful shortcuts in counting.

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

The symbol is an exclamation mark: $n!$ is read as "n factorial". It only makes sense for non-negative integers.

The first few values

$n$	$n!$	Why it matters
0	1	Special case (explained below)
1	1	Only one way to arrange one thing
2	2	Two ways: AB or BA
3	6	Three books, six arrangements
4	24	Four items, 24 arrangements
5	120	Grows fast — five books, 120 ways

Note

$0! = 1$ is not a guess — it's defined to make formulas work. There is exactly one way to arrange zero objects: do nothing. Also, many formulas like $n! = n \times (n-1)!$ would break at $n=1$ if $0!$ weren't 1.

The recursive nature

Factorials have a beautiful pattern: every factorial is the current number times the previous factorial.

5! = 5 \times 4!

4! = 4 \times 3!

3! = 3 \times 2!

2! = 2 \times 1!

1! = 1 \times 0! = 1 \times 1 = 1

This recursive definition is often how you'll compute factorials in problems: $n! = n \times (n-1)!$ , with the base case $0! = 1$ .

Why factorials explode so fast

Notice how quickly the numbers grow: $5! = 120$ , $6! = 720$ , $7! = 5040$ , $10! = 3,628,800$ . By $20!$ , you're already at 2.4 quintillion. This rapid growth is why factorials appear in probability (counting arrangements of decks of cards), combinatorics (choosing teams), and even in advanced mathematics like Taylor series.

Watch out

A common mistake: thinking $n!$ means $n$ multiplied by something else, like $n$ times some number. It's not — it's the product of all integers from $n$ down to 1. Also, factorials are not defined for negative numbers or fractions in basic arithmetic.

The core idea in one sentence

Factorial arithmetic is simply the arithmetic of these products — adding, subtracting, multiplying, and dividing expressions that contain factorials. The key skill is learning to cancel common factors when simplifying, especially in fractions like $\frac{10!}{7!}$ .

For example:

\frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720

You never need to fully expand both factorials — just write out the part that doesn't cancel.

That's the intuition: factorials count arrangements, grow fast, and simplify beautifully when you keep them as products rather than computing the full number.

Factorial Arithmetic is a building block of the NCERT Class 11 Mathematics chapter on Permutations and Combinations, and mastering it is essential before tackling factorial-based important questions in CBSE board exams. Searches like "factorial arithmetic definition, formula and examples" or "n! formula class 11 maths" reflect exactly the kind of foundational practice this concept supports, and it remains a quick-scoring warm-up topic in JEE Main counting problems.

Factorials grow explosively, not additively. Computing both sides shows $3! + 4! = 30$ while $7! = 5040$ , so the equation is false.

Why factorials don't add like ordinary numbers

When you see $3! + 4!$ and wonder if it equals $7!$ , you're testing whether factorials behave like exponents (where $2^3 \cdot 2^4 = 2^7$ ) or like simple addition. They don't. A factorial $n!$ means multiplying all integers from $1$ to $n$ , and this multiplication grows so rapidly that adding two factorials gives a result vastly smaller than the factorial of their sum.

The key insight: $7!$ includes the product $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7$ , which is astronomically larger than just adding $3!$ and $4!$ .

Computing each side

Left side: $3! + 4!$

Start with each factorial:

3! = 3 \times 2 \times 1 = 6

4! = 4 \times 3 \times 2 \times 1 = 24

Adding them:

3! + 4! = 6 + 24 = 30

Right side: $7!$

Now compute the factorial of $7$ :

7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1

Work through the multiplication:

7 \times 6 = 42

42 \times 5 = 210

210 \times 4 = 840

840 \times 3 = 2520

2520 \times 2 = 5040

5040 \times 1 = 5040

So $7! = 5040$ .

Comparison

We have $30$ on the left and $5040$ on the right. These are nowhere close: $7!$ is 168 times larger than $3! + 4!$ .

Watch out

A common misconception is that factorials might combine additively like $a + b = c \implies a! + b! = c!$ . This never holds for $a, b \geq 2$ because factorial growth is multiplicative and explosive.

Tip

To see why factorials grow so fast, notice that $7! = 7 \times 6 \times 5 \times 4!$ . Even if we had $4!$ on the left, multiplying it by $7 \times 6 \times 5 = 210$ to get $7!$ shows the enormous gap.

✓Final answer

3! + 4! \neq 7!

←Previous question · 6.3.2Q1 — Evaluate (i) 8! (ii) 4! - 3!→Next question · 6.3.2Q3 — Compute (8!)/(6!×2!)

Q.Is 3!+4!=7!3! + 4! = 7!3!+4!=7!?

Why factorials don't add like ordinary numbers

Computing each side

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Q.Is $3! + 4! = 7!$ ?

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