Question

12 balls numbered 1 through 12 are placed in a bin. In how many ways can 3 balls be drawn, in order, from the bin, if each ball remains outside the bin after it is drawn?

URGENT!

Answer 1

1320

Explanation:

There are 12 choices for the first ball, 11 remaining choices for the second ball, and 10 remaining choices for the third ball, for a total of 12*11*10=1320 possible drawings.

Answer 2

1320

Explanation:

There are three "spots" for balls.

__ __ __

For the first "spot," and without replacement, meaning we don't put the balls back, there are 12 choices.

12 __ __

But for the next spot, there are only 11 choices. And for the next, there are only 10!

12 11 10

Since these are independent events, we multiply them together.

12 x 11 x 10 = 1320

There are 1320 possibilities!

How could we do it on the calculator, much faster? The question says "in order," so we know it's a permutation, not a combination. This is the nPr button on the calculator (math -> prb -> 2). Just type in 12 nPr 3 and you get the same answer, 1320.