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!
 A: We have the equation
$$
4=k!(4-k)!
$$
for positive integers. Since the only product decomposition is $4=2\cdot 2$ over positive integers, as $4=4\cdot 1=1\cdot 4$ is impossible, it follows $k!=(4-k)!=2$. So $k=2$.
A: As all binomial coefficients, when $k$ increases, $\dbinom 4k$ first increases, then decreases.
To determine the tipping point, consider the ratio of two consecutive coefficients:
$$\frac{\dbinom 4{k+1}}{\dbinom 4k}=\frac{4!}{(k+1)!(4-k-1)!}\frac{k!(4-k)!}{4!}=\frac{4-k}{k+1.}$$
This ratio is $>1$ if & only if $4-k<k+1$, i.e. $k<\frac32$. Therefore, we know that
$$\binom 40 <\binom 4 1 <\binom 42>\binom 43>\binom 44,$$
and it happens that $\dbinom 42=6$ is the tipping point. Therefore, we have single solution: $k=2$.
A: 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, $\boxed{k = 2}$
A: I don't see how to pursue this with any sort of a stepwise procedure that you seem to be asking about.  In general, such procedures succeed mostly when the problem is linear or when the method relies on local linearization (e.g. gradient methods).
I am guessing, you are also thinking of the general class of problems of finding $k$ given a (natural) $N$ and the right-hand side of
$$
\binom{N}{k} = M.
$$
One can try to calculate some bounds on $k$ in your problem:
$$
\binom{4}{k} = {4! \over (4 - k)! k!} = 6,
$$
so
$$
(4 - k)! \; k \leq (4 - k)! \; k! = {4! \over 3} = 8,
$$
etc. (Stirling's approximation is asymptotic, although does pretty well starting with fairly small values of the argument), or rewrite this in terms of the $\Gamma$ function:
$$
\Gamma( 4 - k + 1) \; \Gamma( k + 1) = 8
$$
and see if this can be approached numerically, by temporarily letting $k$ vary continuously.
But, in any case, the problem is nonlinear, and there is no general theory for such problems.  Also, given that the domain of $k$ in your problem is relatively small, the exhaustive search you have used outperforms heavy numerical artillery for your specific problem.
A: 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
A: Just do it.
${4 \choose k} = 6$
$\frac {4!}{k!(4-k)!} = 6$
$k!(4-k)! = \frac {4!}6 = 4$.
So $(k!, (4-k)!)$ equal one of the following $(1,4),(2,2), (4,1)$.
As $2!= 2< 4 < 3!=6$ there it is impossible for either $k!$ or $(4-k)!$ to be equal to $4$.
So $k!=2$ and $(4-k)!= 2$.  And then only $m!=2$ is if $m= 2$.
So $k =2$ and $4-k = 2$.
So $k = 2$.
That's all.
