You take the four Aces, four 2's, and four 3's from a standard deck of 52 cards, forming a set of 12 cards. You then deal all 12 cards at random to four players, so that each player gets three cards. What is the probability that each player gets an Ace, a 2, and a 3?


Can someone help please? Ive been working on this for a few hours and still can't get it...


Any help would be greatly apreciated!

 May 22, 2020



k = 13The smallest zero or root is x = -10


Work Shown:

note: you can write "x^2" to mean "x squared"

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

Hope this helps, whymenotsmart^m^.

 May 22, 2020

umm whymenotsmart, i think you're referring to a different problem?

lokiisnotdead  May 22, 2020

hi guest!


so the first guy can draw 3 cards in \(\dbinom{12}{3}\) ways, and there are 4*4*4 possible draws that contain an ace, a 2, and a 3.

the second guy can draw 3 cards in \(\dbinom{9}{3}\) ways, and there are 3*3*3 possible draws that contain an ace, a 2, and a 3.


the third guy can draw 3 cards in \(\dbinom{6}{3}\) ways, and there are 2*2*2 possible draws that contain an ace, a 2, and a 3.


and, finally, the last guy just gets the remaining cards, which are an ace, a 2, and a 3.


so, the answer is \(\frac{4^3\cdot3^3\cdot2^3}{\binom{12}{3}\binom{9}{3}\binom{6}{3}}=\boxed{\frac{72}{1925}}\)


I hope this helped you!


 May 22, 2020

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