$\big(^n_7\big)$ denotes the number of ways of choosing $7$ objects from among $n$ objects.
For a real number $n , [n]$ denotes the greatest integer not exceeding x.
My attempt :
I simplified $\big(^n_7\big)$ as $\frac {\prod_{j=0}^{6} {(n-j)}} {7!} $.
For ${ \begin{bmatrix}{\frac {n}{7}} \end{bmatrix} }$, the only thing I could infer was that if $n=7k+m $, where $(k,m) \in N$ and $m < 7$ , ${ \begin{bmatrix}{\frac {n}{7}} \end{bmatrix} }$ would equal $k$.
How should I proceed? Can anyone help?
Write $n=7k+i$ where $i\in\{0,1,\ldots,6\}$.
You have: $$\begin{align} \binom{n}{7}&=\frac{(7k+i)(7k+i-1)\cdots(7k+i-6)}{7\cdot6\cdot\cdots\cdot1}\\ &=k\cdot\overbrace{(7k+i)(7k+i-1)\cdots(7k+i-6)}^{\text{remove the factor }7k+i-i}\cdot\frac{1}{6\cdot5\cdot\cdots\cdot1}\\ &\equiv k\mod{7} \end{align}$$
since the remaining six factors in parentheses have residues $\{1,2,\ldots,6\}$, exactly canceling the remaining factors in the denominator when you move to mod $7$ arithemtic.
But also, $$\begin{bmatrix}{\dfrac {n}{7}} \end{bmatrix}=k$$ So $$\binom{n}{7}\equiv\begin{bmatrix}{\dfrac {n}{7}} \end{bmatrix}\mod{7}$$ and therefore $7$ divides their difference.
