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This probably sounds like a really stupid obvious question, but bear with me. I have attached the problem from a lecture from MIT Intro to Probability course. I simply do not understand why multiplying the number of ways to distribute 4 aces by the number of ways to partition the remaining cards is necessary. Can't you just multiply 4C1 with 48C12?

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    Which term are you questioning? – herb steinberg Mar 23 '23 at 22:07
  • To stretch your intuition, consider the situation player by player, starting with the first player to receive cards, Player-1. Assume, without loss of generality, that he is given the first $~13~$ cards off of the top of the deck, and further assume that exactly $~1~$ of these $~13~$ cards is an Ace. How many different card combinations are possible. $$\text{Answer}: ~~\binom{4}{1} \times \binom{48}{12}.$$ Before going any further, does the above computation agree with your intuition? If not, please explain. If so, then what about Player-2? – user2661923 Mar 23 '23 at 23:13
  • Write it as $\dfrac{\dbinom{4}{1,1,1,1}\dbinom{48}{12,12,12,12}}{\dbinom{52}{13,13,13,13}}$ – true blue anil Mar 24 '23 at 04:57
  • @user2661923 Yes, this does agree w/my intuition...I think I understand it now. I was just forgetting some fundamentals and what the multinomial coefficients represents. I understand we need to multiplythe number of ways you can distribute 4 aces with the number of ways to partition the remaining cards. But I'm a bit confused as to how we get from what you provided to the final answer. –  Mar 24 '23 at 19:30
  • So, does the last comment of true blue anil, directly above, now make sense? Do you now understand and agree with the formula presented in your posting? – user2661923 Mar 24 '23 at 19:32
  • @user2661923 I do understand it, yes, thank you for your help. –  Mar 24 '23 at 19:45

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