Solution 1
September 30, 2010
1. For k = 1, it is easy to see that there are n ways to take one thing
out of n things. We assume the number of combinations of n things
n+k1)!
k
taken k at a time is n+k 1 = (k!(n1)! . Now we think about the case
k k + 1. A
Solution 2
October 1, 2010
1. Assume we have n1 steps to the right and n2 steps to the left, so
n1 + n2 = N, n1 n2 = m
N +m
N m
n1 =
, n2 =
.
2
2
(1)
(2)
The probability to have n1 steps to the right and n2 steps to the left is
!
pn1 q n2 , and there are
Solution 3
October 1, 2010
1. (a) From (1.55), with N is very large, we have
= log g = log(N + n 1)! log n! log(N 1)!
log(N + n)! log n! log N !.
(1)
(2)
With the Stirling approximation:
= (N + n) log(N + n) (N + n) (n log n n) (N log N N(3)
)
= (N + n)