A die is rolled 16 times. Given that three of the rolls came up 1, four came up 2, three came up 3, two came up 4, two came up 5, and two came up 6, how many different arrangements of the outcomes are there?

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

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation:

b and c

step-by-step explanation: