Return to Answer

Let $h(x)=-x\ln x-(1-x)\ln (1-x)$ be the binary entropy function in nats, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have $$ \sqrt{\frac{n}{8k(n-k)}}\exp\{-nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{-nh(k/n)\} $$ where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.

Let $h(x)=-x\ln x-(1-x)\ln (1-x)$ be the binary entropy function in nats, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have $$ \sqrt{\frac{n}{8k(n-k)}}\exp\{nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{nh(k/n)\} $$ where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.

This result is most certainly not mine, I learned it from Bob Gallager's Information Theory and Reliable Communications.

Let $h(x)=-x\log_2x-(1-x)\log_2(1-x)$ be the binary entropy function, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have $$ \sqrt{\frac{n}{8k(n-k)}}\exp\{-nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{-nh(k/n)\} $$ where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.

Let $h(x)=-x\ln x-(1-x)\ln (1-x)$ be the binary entropy function in nats, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have $$ \sqrt{\frac{n}{8k(n-k)}}\exp\{-nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{-nh(k/n)\} $$ where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.

Let $h(\cdot)$ be the binary entropy function, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have $$ \sqrt{\frac{n}{8k(n-k)}}\exp\{-nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{-nh(k/n)\} $$ where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.

Let $h(x)=-x\log_2x-(1-x)\log_2(1-x)$ be the binary entropy function, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have $$ \sqrt{\frac{n}{8k(n-k)}}\exp\{-nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{-nh(k/n)\} $$ where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.