In how many ways can a full house of three jacks and a pair of twos be obtained from an ordinary deck of cards?

Answer 1

combination of 4 jacks 3 at a time
C(4,3) = 4! /[ 3! (4-3)! ] = 4

combination of 4 twos 2 at a time
c(4,2) = 4! /[2!(4-2)!] = 3
so
4*3 = 12

Answer 2

c(4,2) = 6 so that the final answer is 4*6 = 24 not 12