The faces of a cube are to be numbered with integers 1 through 6 in such a way that consecutive numbers are always on adjacent faces (not opposite ones). the face numbered 1 is on top and the face numbered 2 is toward the front. in how many different ways can the remaining faces be numbered?

Total different ways to number the remaining faces is 10.

Step-by-step explanation:

It is given that consecutive numbers are always on adjacent faces (not opposite ones).

The face numbered 1 is on top and the face numbered 2 is toward the front.

Top = 1

Front = 2

The remaining faces are left right bottom back. Total possibilities are shown below.

S.no.      3                4              5               6

(1)       Right         Back         Bottom       Left  

(2)       Right         Back         Left         Bottom  

(3)       Right        Bottom       Left           Back  

(4)       Right        Bottom       Back         Left  

(5)       Bottom      Right         Back         Left  

(6)       Bottom       Left          Back        Right  

(7)       Left            Back       Bottom       Right  

(8)       Left            Back        Right         Bottom  

(9)       Left           Bottom      Right          Back  

(10)     Left            Bottom      Back         Right

When Bottom=3, then we can not place 4 on back because doing this 5 and 6 are opposite ones.

Therefore, total different ways to number the remaining faces is 10.

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.