5! == (5 * 4 * 3 * 2 * 1) == 120

int factorial(int n) {

if (n <= 1) {

return 1;

}

else {

return n * factorial(n - 1);

}

}

int factorial(int n) {

return (n <= 1) ? 1 : n * factorial(n - 1);

}

//1) 5 * factorial(5 - 1)

//2) 4 * factorial(4 - 1)

//3) 3 * factorial(3 - 1)

//4) 2 * factorial(2 - 1)

//5) return 1

//So now we have the first difnitive answer

//to return from a function call

//So it goes back down the list with acutal results

//6) return 2 * 1

//7) return 3 * 2

//8) return 4 * 6

//9) return 5 * 24

//So we finally return (5 * 24)

//Wich gives us the answer 120

{1, 3, 4, 5, 8, 10, 12}

/*

Find the middle of the array

Check if the desired number is equal, less,

or greater then the middle number

If it's equal we are done return the middle index.

But it's not it is less than '5'

So we now know its in this section of the array

{1, 3, 4}

So we call the function again within our function

but limit the high and low indecies to 0 and the middle - 1

Now of course the middle number is 3 which is what we wanted!

*/