Correct. The number of ways 10 robots can finish in the top three spots is $\displaystyle\frac{{{10}}}{{{73}}}={120}\usepackage{cmbright}\Huge\$ $\displaystyle{\left({n}={10}{\quad\text{and}\quad}{k}={3}\right)}\usepackage{cmbright}\Huge\$ . The number of ways that these three robots can be grouped is $\displaystyle\frac{{{3}}}{{{\left({3}-{3}\right)}{3}}}={1}\usepackage{cmbright}\Huge\$ $\displaystyle{\left({n}={3}{\quad\text{and}\quad}{k}={3}\right)}\usepackage{cmbright}\Huge\$ since order does not matter. The probability is the ratio of $\displaystyle\frac{{1}}{{120}}\usepackage{cmbright}\Huge\$ . You could also calculate the individual probabilities of each event and multiply: $\displaystyle{\left(\frac{{3}}{{10}}\right)}{\left(\frac{{2}}{{9}}\right)}{\left(\frac{{1}}{{8}}\right)}=\frac{{6}}{{720}}=\frac{{1}}{{120}}\usepackage{cmbright}\Huge\$ .