Probability of a Gin Hand with Cards from Two Suits

[roeb]

Homework Statement

A gin hand consists of 10 cards from a deck of 48 cards (4 Ace Cards are missing).

Find the probability that a gin hand has all 10 cards from two suits.

Homework Equations

The Attempt at a Solution

So my thinking would be since we have 12 cards per suit and we are picking from two out of four possible suits, so Probability = ( 24 pick 10 ) * (4 pick 2) / (48 pick 10 )

In the actual answer there is another factor that I'm missing: (4 pick 2) [ (24 pick 10) - (2 pick 1)*(12 pick 10) ] / ( 48 pick 10)

Does anyone know where this extra term comes from? (2 pick 1)*(12 pick 10)

 

[micromass]

If you write (24 pick 10), then you haven't eliminated the possibility that those 10 cards come from the same suit. I think that they want you to find the probability such that the 10 cards come from two suits (and not all from 1 suit), at least that's how I interpreted the question...