# Is it possible to calculate $\sum_{i=1}^n \ln(x_i!)$, given $\sum_{i=1}^n x_i$

I have a problem solving

$$\sum_{i=1}^n \ln(x_i!)$$

The question gives me only

1. $\sum_{i=1}^n x_i$= $A$

2. Number of samples =$n$

• Hint: $\ln x ! \approx (x+\frac{1}{2})\ln x - x + \ln\sqrt{2\pi}$. – J.G. Aug 10 '17 at 8:28
• If only $A$ is given without the series terms, then no. E.g. $x_1+x_2=1+2=0+3=3=A$ will give $\ln{1!}+\ln{2!}=\ln{2}\ne \ln{0!}+\ln{3!}=\ln{6}$. – farruhota Aug 10 '17 at 11:34
No, we can't find the sum of the logarithms of the factorials from the given information. For instance, if $A = n$, then all the $x_i$ could be $1$, which would make $$\sum_{i = 1}^n\ln(x_i!) = 0$$ or we could have $x_1 = n$ and all the other $x_i = 0$, which would give $$\sum_{i = 1}^n \ln(x_i!) = \ln(n!)$$