Arrangements v. Slot Method

[datadan]

Good Day. I'm making a mistake in one of my methods help me find my error in thinking.

Problem: Find odds of getting 3 aces and 2 kings without replacement.

Method 1 slot method:
(4/52)(3/51)(2/50)(4/49)(3/48)=9x10^(-7)
This is AAAKK, but I can arrange these 10 different ways so after multiplying by 10, I get = 9 x 10^(-6)


Method 2 arrangement method:
My sample space will be 52c5.
My desired selections will be 5c3 (My Aces) and 2c2 (My kings)
so:
[(5c3)(2c2)]/(52c5) = 3.8x10^(-6)

I should get same result but I'm doing something horribly wrong. An assist would be appreciated.

Thank you,

 

[galactus]

There are 4 Aces in a deck and 4 Kings.

\(\displaystyle \frac{\binom{4}{3}\cdot \binom{4}{2}}{\binom{52}{5}}=\frac{1}{108290}\approx .00000923\)

 

[datadan]

Nice.

Thank you.