How to prove if function is increasing Need to prove that function $$P(n,k)=\binom{n}{k}= \frac{n!}{k!(n-k)!}$$ is increasing when $\displaystyle k\leq\frac{n}{2}$.
Is this inductive maths topic?
 A: You’re trying to compare $$\frac{n!}{k!(n-k)!}$$ and $$\frac{n!}{(k+1)!(n-k-1)!}\;.$$ Specifically, you’re trying to show that if $k\le n/2$, then
$$\frac{n!}{k!(n-k)!}<\frac{n!}{(k+1)!(n-k-1)!}\;.\tag{1}$$
Actually, this isn’t quite true: if $n=3$ and $k=1$, the two sides of $(1)$ are equal. $(1)$ is true, however, when $k<\frac12(n-1)$; when $k=\frac12(n-1)$, the two sides of $(1)$ are equal. 
One way to approach this is to show that the difference
$$\frac{n!}{(k+1)!(n-k-1)!}-\frac{n!}{k!(n-k)!}$$
is positive when $k<\frac12(n-1)$ and $0$ when $k=\frac12(n-1)$. This isn’t difficult if you combine the fractions over a common denominator. Remember that $(n-k)!=(n-k)(n-k-1)!$ and $(k+1)!=(k+1)k!$, and you should have little trouble. (Induction is not needed: it’s just algebra.)
A: Write out the quotient $\dfrac{P(n,k+1)}{P(n,k)}$. You get $\dfrac{n-k}{k+1}$. Now the conclusion is obvious.
Induction is not necessary.
A: Actually you can proceed by induction.
Hint Think about the Pascal's triangle: the base cases $n=1$ or $n=2$ are essentially self-evident. Suppose then the thesis to be true for $n$ and say you want it to prove for $n+1$. Then you may want to use the relation $$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$$ to carry out induction. Hope this helps.
A: As $n$ increases, $n!$ increases.  As $n$ increases $(n-k)!$ increases.  $(n+1)!$ is $(n+1)$ times larger than n!.  $(n+1-k)!$ is however only $(n+1-k)$ times greater than $(n-k)!$.  Therefore the numerator increases more than the denominator as n increases.  Therefore P increases as n increases.
As $k$ increases, $k!$ increases.  As $k$ increases $(n-k)!$ decreases.  The question is then, what changes at a greater rate: $k!$ or $(n-k)!$.  Define the functions $A(x) = x!$ and $B(x) = (n-x)!$.  The denominator of $P(n,k)$ is thus $A(k)*B(k)$.
Now $A(k) = k!$ and $B(k) = (n-k)!$.
Also, $A(k+1) = (k+1)! = (k+1)*k!$ 
Also, $B(k+1) = (n - k - 1)! = (n-k)!/(n-k)$. 
Therefore $A(k+1)B(k+1) = (k+1)k!(n-k)!/(n-k) = A(k)B(k)(k+1)/(n-k)$
So $(k+1)/(n-k) \leq 1$ then the function is increasing.
This must be the case since $k \leq n/2$  
Therefore P is an increasing function of $k$ and $n$.
