I want to prove that the binomial coefficient ${n \choose k}$ for $n \ge k$ is a monotonically nondecreasing sequence for a fixed $k$. How do I do this?
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From the recursive formula for binomial coefficient $$ \binom{n}{k} = \binom {n-1}{k} + \binom{n-1}{k-1} \qquad (n, k > 0), $$ it is clear that $\binom{n}{k} \geqslant \binom{n-1}{k}$. The claim is even more obvious when one thinks of the combinatorial interpretation of the binomial coefficient. Every $k$-subset of $\{ 1, 2, \ldots, n-1 \}$ is also a $k$-subset of $\{1, 2, \ldots, n\}$; so it immediately follows that $\binom{n-1}{k} \leqslant \binom{n}{k}$. |
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If if you keep $k$ fixed and increase $n$ the value of $\binom{n}{k}$ will increase monotonically. $$ \begin{align*} \binom{n}{k} &= \frac{n!}{k!(n-k)!} \\ &= \frac{n(n-1)....(n-k+1)}{k!} \tag{1} \end{align*} $$ If you differentiate the RHS of the above relation wrt to $n$ keeping $k$ constant, you will get a positive [or zero derivative: zero derivative occurs for $k=0$] derivative for $n \geqslant k$ [both $n$ and $k$ positive]. Therefore $\binom{n}{k}$ is a monotonocally increasing function of $n$ when $k$ is kept constant. You may consider the function expressed by $(1)$ as a continuous function of $n$ [$k$ is of course a fixed integer]. But our interest will be on the integral values of $n$. These values occupy discrete positions on the domain of continuous function considered. For $k=0$, the function is defined to be a constant ($=1$). |
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Fix $k \geq 0$. Show that $${n+1 \choose k} = \frac{(n+1)!}{k!(n+1-k)!} \geq \frac{n!}{k!(n-k)!} = {n \choose k}.$$ But this follows whenever $n+1 \geq n+1 - k$, so... |
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