Let $k=n+1$, so that $n=k-1$, and substitute into $$\sum_{n\ge 0}\binom{n+2}2z^{n+1}$$ to get $$\sum_{k-1\ge 0}\binom{(k-1)+2}2z^k=\sum_{k\ge 1}\binom{k+1}2z^k\;;$$ then simply rename $k$ back to $n$ to get
$$\sum_{n\ge 1}\binom{n+1}2z^n\;.$$
(Note that it’s now $n+1$ in the binomial coefficient, not $n+2$.) Finally,
$$\binom{n+1}2=\frac{(n+1)!}{2!\big((n+1)-2\big)!}=\frac{(n+1)!}{2(n-1)!}=\frac{n(n+1)}2\;.$$
Here is a copy of an answer I gave to a similar post (I'll edit it to fit your question a bit better):
The formula for permutations is $\frac{n!}{(n-k)!}$
The formula for combinations is $\frac{n!}{k!(n-k)!}$
(Here I assume you understand the permutation formula a bit). To derive the second from the first let's make an example... say we are finding the combinations of $3$ numbers from a set of $4$, say $1,2,3,$ and $4$. If we use the permutation formula we get our answer as $24$, but we can see we over-counted... for example, $1,2,3$ and $1,3,2$ were counted separately. Since we are taking sets of size $3$ and there are $3!$ ways to arrange a set of size $3$, we have over-counted by a factor of $3!$. Generalizing this, if we use the permutation formula to get combinations we will always over-count by $k!$, where $k$ is the size of the set we take from the whole, and hence the formula for combinations above (we say order does not matter with combinations because of that division... every set we take is unique no matter how we arrange it. This is not true for permutations).
EDIT: Since you are new to this, we get the permutations formula by first choosing one from a set of $n$, then $(n-1)$, then $(n-2)$, and we do this until we get $(n-k+1)$, yielding $P(n,k)=n(n-1)(n-2)(n-3)\ldots(n-k+1)$ (The last case avoids $0$ when $n=k$). We then multiply by (n-k)! to get $$P(n,k)=\frac{n(n-1)(n-2)(n-3)\ldots(n-k+1)(n-r)!}{(n-r)!}$$ $$= \frac{[n(n-1)(n-2)\ldots(n-k+1)]\cdot[(n-k)(n-k-1)\ldots(3)(2)(1)]}{((n-r)!}$$ $$= \frac{n!}{(n-r)!}$$
(The last step requires an observation about the descending nature of the numerator... I leave it as a math puzzle to you!)
Best Answer
$$\frac{(x-2)!}{(x-4)!}-46=\frac{x!}{(x-1)!}\iff \frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46=0$$ $$\frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46= \frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}\cdot \frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46= $$ $$=\frac{[(x-2)(x-3)]\cdot(x-1)!}{(x-1)!}-\frac{x\cdot(x-1)!}{(x-1)!}-\frac{46(x-1)!}{(x-1)!}=(x-2)(x-3)-x-46 $$