Simple Blackjack Probability Question

Let's say you're playing SD blackjack and playing 5 hands at once....

What is the probability of being dealt exactly ONE TEN to those five hands against a dealer ACE...

My guess:

4 (dealer aces) x 16 (player tens) x 35 (remaining non-tens to player) x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 10 (There are 10 different positions to deal a 10 to the player)
/ 52 (cards remaining) x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42

= .0068 = .68%

Thanks,

-railer

That's correct: (4/52) * 16*C(35,9)/C(51,10) = 0.68%

Another approach is, instead of choosing cards for the player, choose places for the tens:
(4/52) * 10*C(41,15)/C(51,16) = 0.68%

Cheers