We sometimes need to bound the size of a binomial coefficient. For 1 <= k= < n , we have the lower bound as
(n
 k)     = n ( n-1) ... (n-k + 1)   /   k.k-1...1
which we can rearrange to get >= (n/k) ^ n
Here use the inequality k! >= (k / e ) ^ k  derived from Stirling's approximation, we obtain the upper bounds
(n
k)     = n.(n-1)...(n-k+1) / k. k-1. ... 1    <= (n^k / k!)   <= (en/k) ^ k
for all 0 =k <= n , we can use mathematical induction to prove the bound
(n
k)   <=  n^n  /  k^k.(n-k)^(n-k)
where for convenience we assume that 0 ^ 0 = 1
i.e for the trivial case we have k = 0, we have
(n
0) <= 1
and for k we assume  that it holds, now we look at k+1
(n
k+1) <= n^n/ (k+1)^(k+1).(n-k-1)^(n-k-1) which we rearrange to
write as <= n^n / k^k.(n-k)^(n-k) because the denominator increases with k+1
and it holds for k+1
For k = lambda.n, where 0<=lambda<=1, this bound can be rewritten as
(n
lambda.n) <= n^n / (lambda.n) ^ (lambda.n) . (((1-lambda)n) ^ ((1-lambda)n))