Ever looked for a simple explanation for: why 0!=1 ?

I am sure your teachers may have explained you through this:

n! = (n-1)!n

hence, 1! = 0!1

dividing both sides by 1, 1! / 1 = 0!1 / 1

hence 1 = 0!

But here I am going to explain in a different way.

Look at this set: {a,b,c}. It has 3 elements: a , b , and c . Now arrange all these elements in different ways, and let us see how many different arrangements are possible.

Arrangement 1: {a,b,c}

I am sure your teachers may have explained you through this:

n! = (n-1)!n

hence, 1! = 0!1

dividing both sides by 1, 1! / 1 = 0!1 / 1

hence 1 = 0!

But here I am going to explain in a different way.

Look at this set: {a,b,c}. It has 3 elements: a , b , and c . Now arrange all these elements in different ways, and let us see how many different arrangements are possible.

Arrangement 1: {a,b,c}

Arrangement 2: {b,c,a}

Arrangement 3: {c,a,b}

Arrangement 4: {a,c,b}

Arrangement 5: {c,b,a}

Arrangement 6: {b,a,c}

So with a set of 3 elements, 6 arrangements are possible. So we have 3! = 6.

Similarly this set: {a,b}, with 2 elements has 2 possible arrangements: {a,b} and {b,a}. That makes 2!=2.

Similarly this set: {a}, with 1 element has 1 possible arrangement: {a}. That makes 1!=1.

Now look at this set, it has no element: {}.

In how many ways can you arrange it's elements ? The answer is, there is only 1 possible arrangement: {}.

So that brings us to this result: 0!=1

So with a set of 3 elements, 6 arrangements are possible. So we have 3! = 6.

Similarly this set: {a,b}, with 2 elements has 2 possible arrangements: {a,b} and {b,a}. That makes 2!=2.

Now look at this set, it has no element: {}.

In how many ways can you arrange it's elements ? The answer is, there is only 1 possible arrangement: {}.

So that brings us to this result: 0!=1