What is the probability that Bo, Colleen, Jeff, and Rohini win the first, second

Definitions
Definition permutation (importance of order):
No repetition allowed: ${\displaystyle P(n,r)=\frac{n!}{(n-r)!}}$
Repetition allowed ${\displaystyle n^r}$
Combination definition (order is not crucial):
No repetition allowed: $C(n,r)=\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}$
Repetition allowed: ${\displaystyle C(n+r-1,r)=\frac{(n+r-1)!}{r!(n-1)!}}$
with ${\displaystyle n\ne n\cdot (n-1)\cdot \dots \cdot 2\cdot 1}$
Solution:
Numerous methods for choosing the winners
Since the first, second, and third prizes are different, the order of the winners is important, and we must thus utilize the definition of a permutation.
Out of the 50 participants, 4 winners must be chosen.
${\displaystyle n=50}$
${\displaystyle r=4}$
Since repetition of winners is not allowed:
${\displaystyle P(50,4)=\frac{50!}{(50-4)!}=\frac{50!}{46!}=50\cdot 49\cdot 48\cdot 47=5,527,200}$
Probability
Bo, Collen, Jeff, and Rohini will each receive the first, second, third, and fourth prizes in 1 of the 5,527,200 possible combinations.
The probability is calculated by dividing the number of likely outcomes by the total number of outcomes.
$P(Bo,Colleen,JeffandRohini)=\frac{\text{\# of favorable outcomes}}{\text{\# of possible outcomes}}=\frac{1}{5,527,200}\approx 0.000001809=1.809\times {10}^{-7}$
b) Number of methods used to choose the winners
We require the concept of a permutation since the order of the winners matters (because the first, second, and third prizes are different).
Out of the 50 participants, 4 winners must be chosen.
${\displaystyle n=50}$
${\displaystyle r=4}$
Since winners may be selected more than once:
${\displaystyle n^r={50}^4=6,250,000}$
Probability
Bo, Collen, Jeff, and Rohini will each receive the first, second, third, and fourth prizes in 1 of the 6,250,000 possible combinations.
The probability is calculated by dividing the number of likely outcomes by the total number of outcomes.
$P(Bo,Collen,JeffandRohini)=\frac{\text{\# of favorable outcomes}}{\text{\# of possible outcomes}}=\frac{1}{6,250,000}\approx 0.00000016=1.6\times {10}^{-7}$
Result: a) ${\displaystyle \frac{1}{5,527,200}}$
b) ${\displaystyle \frac{1}{6,250,000}}$