Need help with Math probability question

For each of k = 0,1,2,3,4 find the probability that a poker hand( five cards) contains just k aces.

Thanks

Lets say C(n,k) be the combination of n and k which is equal to n! / (k! (n-k)!), we have
k=0
C(4,0)*C(48,5)/C(52,5)= (1*1712304) / 2598960= 0.658841998

k=1
C(4,1)*C(48,4)/C(52,5)= (4*194580) / 2598960= 0.299473636

k=2
C(4,2)*C(48,3)/C(52,5)= (6*17296) / 2598960= 0.039929818

k=3
C(4,3)*C(48,2)/C(52,5)= (4*1128) / 2598960= 0.001736079

k=4
C(4,4)*C(48,1)/C(52,5)= (1*48) / 2598960= 1.84689E-05