Alock has 60 digits, and the combination involves turning either left or right to the first numbers, turning either left or right to the second number, and turning either left or right to the third number. how many possible combinations are there?

Since there are a total of 60 digits, therefore this means that there are 60 digits on the right and then when we turn left we still have 60 digits since the lock forms a circle.

There are two ways to do the combination, either turn left first or turn right first.

Just for calculation, let us assume we turn left first and assume that the numbers can be repeated.

So if we turn left first, we can see that we have 60 digits to choose from. Then the next step is to turn right, and we see that we still have 60 digits to choose from. Now the last turn is to turn left again, and we still have 60 digits to choose from.

Therefore the total combinations would be = 60 * 60 * 60 = 216,000

Since this value is only for turning left first, we multiply this by 2 since we can also turn right first.

Therefore the total possible combinations are:

possible combinations = 216,000 * 2

possible combinations = 432,000 ways

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

it is impossible.