Answer: 1) 1/3,654 2) 3/406 3) 72,684,900,288,000 4) 120
Step-by-step explanation:
1) First and Second and Third
[tex]\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}[/tex]
2) First and Second
[tex]\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}[/tex]
[tex]3)\quad \dfrac{29!}{(29-10)!}=\large\boxed{72,684,900,288,000}[/tex]
[tex]4)\quad _{10}C_3=\dfrac{10!}{3!(10-3)!}=\large\boxed{120}[/tex]