+0

0
32
5

https://web2.0calc.com/questions/help_71935

Sadly all the answers I got were wrong... Can someone help? THANKS!

May 22, 2020

#1
+1498
+3

You said you've been working on it, in a few hours you should make a bit of progress, can you share what you have done so far?

May 22, 2020
#2
0

I tried to calculate the number of possibilities of Aces, 2s and 3 in the same hand as (4^3\times3^3\times2^3\times1^3)=13824, and the number of possibilities as \dbinom{12}{3}\dbinom{8}{3}\dbinom{4}{3}=49280. 13824/49280=\boxed{\frac{108}{385}}

Sorry, the latex doesn't work here...

Guest May 22, 2020
#3
+438
+1

good job! you're on the right track! the numerator is correct, but the denominator isn't.

you got the 12 choose 3 part right, but not the rest.

think about it, if you've already chosen 3 cards, then there would be 12-3=9 cards left to choose from, not 8.

you're really close! you got this!

:)))

#4
+1

Wait. So I should have done 12 choose 3 times 9 choose 3 time 6 choose 3?

May 22, 2020
#5
+438
+1

yes!!!