SOLUTION: I submitted this problem Friday evening and I got no response. I need to know if I'm right or wrong, it's due tomorrow. If 10 cards are drawn without replacement from an ordinary

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Question 15532: I submitted this problem Friday evening and I got no response. I need to know if I'm right or wrong, it's due tomorrow.
If 10 cards are drawn without replacement from an ordinary deck, find the probability of obtaining exactly 3 aces or exactly 3 kings (or both). I tried
using (52 choose 10) for the denominator and (10 choose 3)^4 for the numerator(4 for each suite). It seems as though I'm missing something. Please help asap.

Found 2 solutions by venugopalramana, biplavmehta:
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!

If 10 cards are drawn without replacement from an ordinary deck, find the probability of obtaining exactly 3 aces or exactly 3 kings (or both). I tried
using (52 choose 10) for the denominator and (10 choose 3)^4 for the numerator(4 for each suite). It seems as though I'm missing something. Please help asap.
Total number of possibilities. N = ...taking 10 out of 52 cards = 52C10
number of successes....M =...taking 3 aces and rest seven other than aces.
taking 3 aces out of 4 aces ..=4C3
taking 7 cards out of rest other than aces...=48C7
M = 4C3 * 48C7
probability = M/N = 4C3 * 48C7 /52C10 = (4*48*47*46*45*44*43*42)*(10*9*8*7*6*5*4*3*2*1)/(7*6*5*4*3*2*1)*(52*51*50*49*48*47*46*45*44*43)
FOr 3 kings only the answer is same
for both 3 kings and 3 aces you can try your self on the above basis

Answer by biplavmehta(3) (Show Source):
You can put this solution on YOUR website!
total no. of cases= 52c10
if 3ace are selected favorable no. of cases= 4c3*48c7
if 3kings are selected favorable no. of cases= 4c3*48c7
if 3ace and 3kings selected favorable no. of cases= 4c3*4c3*46c4
final result= (4c3*48c7 + 4c3*48c7 + 4c3*4c3*46c4) / 52c10