Answer :
Answer:
there are 3.027*10^9 different arrangements
Step-by-step explanation:
if we go one group of outcomes at the time then,
arrangements = possible arrangements with the first group * possible arrangements with the second group given that the fist group was already allocated *possible arrangements with the second group given that the first and second group were already allocated ...
thus
arrangements = N!/[(N-n₁)!*n₁!] * (N-n₁)!/[(N-n₁-n₂)!*n₂!] * (N-n₁-n₂)!/[(N-n₁-n₂-n₃)!*n₃!] ....
thus the number of arrangements is
arrangements = N! /(n₁!*n₂!*n₃!*n₄!*n₅!*n₆!)
where
N = total number of times the dice is rolled
n₁,₂...₆ = number of times the dice outcome is 1,2 ... 6 respectively
replacing values
arrangements = 16!/(3!*4!*3!*2!*2!*2!) = 3.027*10^9 different arrangements