<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] You can write $n\choose{k}$ = $\frac{n!}{(n-k)!k!}$. Substituting $n = 4$ into there, we get $\frac{24}{(4-k)!k!}$ = 6. Multiplying both sides by $(4-k)!k!$, we get $24 = 6(4-k)!(k!)$. Dividing both sides by $6$, we get $4 = (4 - k)!(k!)$. Realizing $4 = 2^2$, we can assume that $k! = 2$ and $4 - k = 2$.
The only value of $k$ such that $k! = 2$ is $2$.
Hence, $\boxed{k = 2}$
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