Correct. The terms $\displaystyle{16}{{m}}^{{2}}{n}\usepackage{cmbright}\Huge\$ and $\displaystyle-{5}{n}{{m}}^{{2}}\usepackage{cmbright}\Huge\$ are like terms. They can be combined this way: $\displaystyle{16}{{m}}^{{2}}{n}-{5}{n}{{m}}^{{2}}={16}{{m}}^{{2}}{n}-{5}{{m}}^{{2}}{n}={\left({16}-{5}\right)}{{m}}^{{2}}{n}={11}{{m}}^{{2}}{n}\usepackage{cmbright}\Huge\$