Can someone help me solve this question? I've read the book, notes, I still don't get it...
Simply each of the following.
a) n!/(n-1)!
b) (n-r)!/(n-r-1)!
think of numbers, eg 6!
what is 6!/5! ?
= 6x5x4x3x2x1/(5x4x3x2x1)
= 6
same way for n!/(n-1)!
= n(n-1)(n-2)..(3)(2)(1) / [(n-1)(n-2)..(3)(2)(1)]
= n
try the other problem
I don't understand it...
n(n-r)(n-r-1)(n-r-2)..(3)(2)(1)/(n-r-1)(n-r-2)..(3)(2)(1)
why did you start with n! ?
it said:
(n-r)!/(n-r-1)!
which is
(n-r)(n-r-1)(n-r-2) .. (3)(2)(1) /((n-r-1)(n-r-2)...(3)(2)(1)
= n-r
Oh ok, thank you! I kind of understand it now and will do some extra questions.
Of course, I'd be happy to help you understand these questions!
To solve these expressions, we'll start by understanding the concept of factorials. The exclamation mark (!) denotes a factorial.
The factorial of a number is the product of that number and all positive integers less than it. In other words, n! is equal to n multiplied by (n-1) multiplied by (n-2), and so on, down to 1.
Now, let's solve the expressions step by step.
a) n!/(n-1)!
To simplify this expression, we can cancel out the common factor of (n-1)! in both the numerator and denominator.
So, n! divided by (n-1)! is equal to n multiplied by (n-1)!, divided by (n-1)!.
Since (n-1)! cancels out, we are left with n.
Therefore, the answer to expression a) is n.
b) (n-r)!/(n-r-1)!
Similar to the previous expression, we can cancel out the common factor of (n-r-1)!.
So, (n-r)! divided by (n-r-1)! is equal to (n-r) multiplied by (n-r-1)!, divided by (n-r-1)!.
Again, (n-r-1)! cancels out, leaving us with (n-r) as the answer.
Therefore, the answer to expression b) is (n-r).
In summary:
a) n!/(n-1)! simplifies to n.
b) (n-r)!/(n-r-1)! simplifies to (n-r).
I hope this explanation helps you understand these expressions better! If you have any further questions, feel free to ask.