333 
of Edinburgh, Session 1873 - 74 . 
Similarly, for the probability that all the particles of nitrogen are 
in the space JB, we find 
/ 6 \2x 10 12 
UTb) 
Hence the probability that all the oxygen is in A and all the 
nitrogen in B is 
/ a \2 x 10 12 / b \8 x 10 12 
\a + b) \a + b) 
Now by hypothesis 
and therefore 
a 2 
cTT~b = 10 ’ 
= . 
a + b 10 ’ 
hence the required probability is 
0 26 x 10 12 
10 
io u 
Call this — , and let log denote common logarithm. 
We have 
log N = 10 13 - 26 x 10 12 x log 2 = (10 -26 log 2) x 10 12 = 2173220 
x 10 6 . This is equivalent to the result stated in the text above. 
The logarithm of so great a number, unless given to more than 
thirteen significant places, cannot indicate more than the number of 
places of whole numbers in answer to the proposed question, ex- 
pressed according to the Arabic notation. 
The calculation of T 4 , when i and n-i are very large numbers, 
is practicable by Stirling’s theorem, according to w T hioh we have 
approximately 
1.2. • 
‘A, 
and therefore 
n(n — 1 ) . . . . bn — % + 1 ) n n + i 
1.2 .... i J27ri i+ *(n-i) n ~ i + lt 
