Here is a much easier way to solve this. The method does not use Hypergeometric selection
There are \((3^3) = 27\) arrangements of success. Divide the number of successes by the total number of sets giving \((3^3) / (nCr(9,3) = \dfrac {9}{28} \approx 32.14\%\)
This matches your calculation, Catmg.
I’m glad you are paying attention to my presentations and train wrecks! 
If my train was loaded with toxic chemicals, I could wipe out a city. Or, in this case, at least send a student down the wrong track.
The calculation is exactly twice the value I presented for the Hypergeometric solution.
The formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...
I saved the input used for the calculation: (3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))
...I used a three (3) instead of a four (4) in the second binomial.
[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.] 
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Here’s a graphic, demonstrating the arrangements for the above solution.
\(\hspace {.3em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.3em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ X & X & X \\ X & X & X \\ \end{array} } \right] \text {First arrangement of “R} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & R \\ X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & X \\ X & X & R \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & X & R \\ X & R & X \\ X & X & X \\ \end{array} } \right] \text {Fourth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .} \\ \, \\ \left[ {\begin{array}{ccc} X & X & X \\ X & X & X \\ R & R & R \\ \end{array} } \right] \text {27^{th} arrangement} \\ \)
From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.
GA
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