E t e r n i t y e n d l e s s
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
n p =n!/(n-r)! r and n c =n!/r!(n-r)! r
This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
Combinations of r from n without replacement is c(n,r) = n!/(n-r)!r! c(n,r) = 23!/20!3! c(n,r) = 1771.
nCr=n!/(r!(n-r)!)
n(n-r)/r
P(n,r)=(n!)/(r!(n-r)!)This would give you the number of possible permutations.n factorial over r factorial times n minus r factorial
If you have n objects and you are choosing r of them, then there are nCr combinations. This is equal to n!/( r! * (n-r)! ).
julian r. geiger
The coefficient of x^r in the binomial expansion of (ax + b)^n isnCr * a^r * b^(n-r)where nCr = n!/[r!*(n-r)!]
nCr=n!/r!/(n-r)!