<s>[INST] The equation ${4 \choose k}=6$ Find the solution of the equation $${4 \choose k}=6.$$
So $k$ must be a natural number $(n\in \mathbb{N})$ and we can find that when $k=1 \rightarrow {4\choose 1}=4$ and when $k=2\rightarrow {4\choose2}=6$, so the solution of the equation is actually $k=2$. Can we solve the problem without calculating the exact value of ${4\choose k}$? I tried: $$\dfrac{V_4^k}{k!}=6$$ but don't see how to solve this. Thank you in advance!
 [/INST] For this case, you can use Pascal's triangle and binomial expansion.
$$
Pascal's\ \ triangle:
\\
\newcommand\pad[1]{\rlap{#1}\phantom{50}}
\begin{matrix}
n=0:&&&&&&&1\\
n=1:&&&&&&1&&1\\
n=2:&&&&&1&&2&&1\\
n=3:&&&&1&&3&&3&&1\\
n=4:&&&\pad{1}&&\pad{4}&&\pad{6}&&\pad{4}&&\pad{1}\\
n=5:&&\pad{1}&&\pad{5}&&\pad{10}&&\pad{10}&&\pad{5}&&\pad{1}
\end{matrix}
$$
$$
binomial\ \ expansion:
\\
\sum_{k=0}^n \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n-1} + \binom{n}{n} =  2^n = (1 + 1)^n $$
In your case,
$$ (1 + 1)^4 = 1 + 4 + 6 + 4 + 1 $$
So $ k = 2 $.
For example, for $ (1 + 1)^2 = 1 + 2 + 1 $:
$$  \binom{2}{0} = 1 , \binom{2}{1} = 2,  \binom{2}{2} = 1 $$
More information about Pascal's triangle and binomial expansion: https://en.wikipedia.org/wiki/Pascal's_triangle
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