Hi guys!
I really need to solve this expression without the help of the computer but only using a simple calculator...please help!!:wink:
(4965!*4950!)/(4915!*5000!)
I've also tried to use the > > > Stirling's approximation < < < but ...
Here is a method using Stirling's approximation:
\(\displaystyle ln(x) \ \ stands \ \ for \ \ log_e(x), \ \ \) the natural logarithm of x
\(\displaystyle ln(n!) \ \ is \ \ asymptotic \ \ to \ \ \ n*ln(n) \ - \ n.\)
Let K = (4965!*4950!)/(4915!*5000!)
Let A = 4965
Let B = 4950
Let C = 4915
Let D = 5000
\(\displaystyle Then \ \ K \ \ = \ \dfrac{(A!)*(B!)}{(C!)*(D!)}\)
\(\displaystyle ln(K) \ = \ ln(A!) \ + \ ln(B!) \ - \ ln(C!) \ - ln(D!)\)
\(\displaystyle ln(K) \ \approx \ [A*ln(A) \ - \ A] \ + \ [B*ln(B) \ - \ B] \ - \ [C*ln(C) \ - C ] \ - \ [D*ln(D) \ - \ D]\)
\(\displaystyle ln(K) \ \approx \ A*ln(A) \ - \ A \ + \ B*ln(B) \ - \ B \ - \ C*ln(C) \ + C \ - \ D*ln(D) \ + \ D\)
\(\displaystyle ln(K) \ \approx \ A*ln(A) \ + \ B*ln(B) \ - \ C*ln(C) \ - \ D*ln(D) \ - \ A \ - \ B \ + \ C \ + \ D\)
\(\displaystyle - A \ - \ B \ + \ C \ + \ D \ = \ -4965 \ - \ 4950 \ + \ 4915 \ + \ 5000 \ = \ 0\)
\(\displaystyle ln(K) \ \approx \ 4965*ln(4965) \ + \ 4950*ln(4950) \ - \ 4915*ln(4915) \ - \ 5000*ln(5000)\)
\(\displaystyle ln(K) \ \approx \ -0.353\)
\(\displaystyle e^{ln(K)} \ \approx \ e^{-0.353}\)
\(\displaystyle K \ \approx \ 0.70\)