Answer
$4845a^{16}b^{16}$
Work Step by Step
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!},\qquad n!=n(n-1)\cdot...\cdot 2\cdot 1,\ \ n!=0$
--------
$...$In all the terms, exponents of a and b add to n...
In $(a+b)^{20} $,the fifth term in the expansion is where
the exponent of b is $5-1=4,$
the exponent of a is $20-4=16,$ and
the coefficient is $\displaystyle \left(\begin{array}{l}
20\\
4
\end{array}\right)=\frac{20\cdot 19\cdot 18\cdot 17}{1\cdot 2\cdot 3\cdot 4}=4845$
The fifth term is: $ 4845a^{16}b^{4}.$
So, for $(ab-1)^{20}$ this term is
$4845(ab)^{16}(-1)^{4}=+4845a^{16}b^{16}$
