The sum of the series $$2 \times 1 \times {^{20}C_4} - 3 \times 2 \times {^{20}C_5} + 4 \times 3 \times {^{20}C_6} - 5 \times 4 \times {^{20}C_7} + \ldots + 18 \times 17 \times {^{20}C_{20}}$$, is equal to

Write a general term for the given series.
For the term containing $${}^{20}C_r$$ (where $$4 \le r \le 20$$) we have

$$\text{term} = (-1)^r\,(r-2)(r-3)\,{}^{20}C_r$$
because the sign alternates starting with $$+$$ for $$r=4$$, and
$$(r-2)(r-3)=2 \cdot 1,\; 3 \cdot 2,\; 4 \cdot 3,\ldots$$ exactly as in the question.

Denote the required sum by $$S$$:

$$S=\sum_{r=4}^{20}(-1)^r\,(r-2)(r-3)\,{}^{20}C_r$$ $$-(1)$$

First evaluate the easier sum running from $$r=0$$ to $$20$$ and then subtract the unwanted first four terms.

Define

$$T=\sum_{r=0}^{20}(-1)^r\,(r-2)(r-3)\,{}^{20}C_r$$ $$-(2)$$

Expand $$(r-2)(r-3)$$ so that each part can be handled by standard binomial identities:

$$(r-2)(r-3)=r(r-1)-5r+6$$

Therefore

$$T=\sum_{r=0}^{20}(-1)^r r(r-1)\,{}^{20}C_r -5\sum_{r=0}^{20}(-1)^r r\,{}^{20}C_r +6\sum_{r=0}^{20}(-1)^r {}^{20}C_r$$ $$-(3)$$

Each of the three sums in $$(3)$$ can be evaluated by differentiating $$(1+x)^{20}$$ at $$x=-1$$.

1. $$\displaystyle \sum_{r=0}^{20}(-1)^r{}^{20}C_r=(1-1)^{20}=0$$

2. Differentiate once:
$$20(1+x)^{19}=\sum_{r=0}^{20}r\,{}^{20}C_r x^{\,r-1}$$
Multiply by $$x$$ and put $$x=-1$$:

$$\sum_{r=0}^{20}(-1)^r r\,{}^{20}C_r = -20(1-1)^{19}=0$$

3. Differentiate twice:
$$20\!\cdot\!19\,(1+x)^{18}=\sum_{r=0}^{20} r(r-1)\,{}^{20}C_r x^{\,r-2}$$
Multiply by $$x^2$$ and put $$x=-1$$:

$$\sum_{r=0}^{20}(-1)^r r(r-1)\,{}^{20}C_r = 20\!\cdot\!19\,(1-1)^{18}=0$$

Hence every sum in $$(3)$$ equals $$0$$, so

$$T=0$$ $$-(4)$$

We now remove the terms for $$r=0,1,2,3$$ to obtain $$S$$.

Compute the four omitted terms:

For $$r=0$$: $$(-1)^0(0-2)(0-3){}^{20}C_0 = 6$$

For $$r=1$$: $$(-1)^1(1-2)(1-3){}^{20}C_1 = -2 \times 20 = -40$$

For $$r=2$$: $$(-1)^2(0)(-1){}^{20}C_2 = 0$$

For $$r=3$$: $$(-1)^3(1)(0){}^{20}C_3 = 0$$

The total of the excluded terms is $$6-40=-34$$.

Because $$T=0$$ from $$(4)$$, we have

$$S = T - (\text{excluded terms}) = 0 - (-34) = 34$$

Thus, the value of the given series is $$34$$.