where can i find the derivation of the permutation formula and the proving that 0 factorial is equal to one?

Answer 1

the number of ways in which r places can be filled is
n(n-1)(n-2)(n-3)................. factors
nPr=n(n-1)(n-2).........(n-(r-...
that is nPr=n(n-1)(n-2)......(n-r+1)
=>n(n-1)(n-2).....(n-r+1)(n-r)...
(n-r)(n-r-1).........1
so nPr=n!/(n-r)!
so nPn=n!
nP0=n!/(n-0)!=n!/n!=1
if r=n,nPn=n!/(n-0)!=n!/0!
But nPn=n!
so n!/0!=n!
=>0!=1

Answer 2

x^0=a million bcuz x^0= x^(n-n)=x^n/x^n=a million!!!!!!!!! Why does 0! = a million ? commonly n factorial is defined interior here way: n! = a million*2*3*...*n yet this definition does not supply a cost for 0 factorial, so a organic question is: what's the fee right here of 0! ? a generic thank you to work out that 0! = a million is by ability of working backward. all of us be attentive to that: a million! = a million 2! = a million!*2 2! = 2 3! = 2!*3 3! = 6 4! = 3!*4 4! = 24 we are able to turn this around: 4! = 24 3! = 4!/4 3! = 6 2! = 3!/3 2! = 2 a million! = 2!/2 a million! = a million 0! = a million!/a million 0! = a million in this way a existence like fee for 0! could be got here across. How can we in fantastic condition 0! = a million right into a definition for n! ? enable's rewrite the final definition with recurrence: a million! = a million n! = n*(n-a million)! for n > a million Now that's difficulty-free to alter the definition to comprise 0! : 0! = a million n! = n*(n-a million)! for n > 0 Why is it important to compute 0! ? an important application of factorials is the computation of style mixtures: n! C(n,ok) = -------- ok!(n-ok)! C(n,ok) is the style of mixtures you could make of ok gadgets out of a given set of n gadgets. We see that C(n,0) and C(n,n) would desire to be equivalent to a million, yet they require that 0! be used. n! C(n,0) = C(n,n) = ---- n!0! So 0! = a million smartly fits what we anticipate C(n,0) and C(n,n) to be

Answer 3

The value of factorial zero is based on something called 'gamma function' which you will learn in higher classes. There is a formula by which you can find the gamma function of every real number. If the number is an integer, then it gives the value of (n-1)!.

Gamma of 1 is 1. Hence in mathematical formulae it was logical to take 0! as 1.

Answer 4

Go to your college library and search few books, look at the chapter of interest. You will find it in one of them.

0! = 1
Hint.
0! can be express as "n" to power Zero so any thing to power zero is 1

Answer 5

the number of ways in which r places can be filled is
n(n-1)(n-2)(n-3)....................r factors
nPr=n(n-1)(n-2).........(n-(r-1))
that is nPr=n(n-1)(n-2)......(n-r+1)
=>n(n-1)(n-2).....(n-r+1)(n-r)(n-r-1)........1/
(n-r)(n-r-1).........1
so nPr=n!/(n-r)!
so nPn=n!
nP0=n!/(n-0)!=n!/n!=1
if r=n,nPn=n!/(n-0)!=n!/0!
But nPn=n!
so n!/0!=n!
=>0!=1