5 handed, 5 pocket pairs dealt-- odds of no flopped sets?

[djk123]

If in a 5 handed game 5 different pocket pairs were dealt, would this be the way to go about calculating the odds of no one flopping a set?

( nCr(10,0) * nCr(32,2) )/ ( nCr(42,3) )

which comes out to equal : 4.3206%

[BruceZ]

[ QUOTE ]
If in a 5 handed game 5 different pocket pairs were dealt, would this be the way to go about calculating the odds of no one flopping a set?

( nCr(10,0) * nCr(32,2) )/ ( nCr(42,3) )

which comes out to equal : 4.3206%

[/ QUOTE ]

C(32,3) / C(42,3) =~ 42.3%

[djk123]

o crap, i do not know why i did (32,2). thanks, stupid error on my part.