Question

4. Tina has 12 balls. Every ball is a different color. She randomly lines the balls up next to each other. How many different ways can the 12 balls be lined up?


A. 13!

B. 11!

C. 10!

D. 12!

Answer 1

How many choices does she have for the first ball to put in line?

Answer 2

12

Answer 3

Precisely. Now that she's already chosen one, how many options for the second ball?

Answer 4

11

Answer 5

We're doing good. If we continue on this pattern, what's your guess to the answer to the original question?

Answer 6

Perhaps I should ask how many choices Tina has for the first two places first. If she has 12 choices for the first ball, and 11 for the second, how many choices would she have in total for the first two spots?

Answer 7

12

Answer 8

Not quite. We have to multiply the options together. So we should get \(12\cdot11=132\). For each ball we put down first, we can choose a certain number of balls to put down second, and by multiplying we get the actual amount we need.

Answer 9

To help you see this, suppose we have 3 balls. Red, green, blue. We can choose 3 balls to begin with and 2 balls for the second one in line. So for the first two balls we have the options
(Red, Green)
(Red, Blue)
(Green, Red)
(Green, Blue)
(Blue, Red)
(Blue, Green)

Answer 10

There are three choices each time for the first ball, and for each choice we have two more choices for the second ball. Hence, we have \(3\cdot2=6\) total choices.

Answer 11

So now, if we return to the original problem, how many choices does Tina have for the first two positions if she has 12 choices for the first, and 11 for the second?

Answer 12

and 10 for the third

Answer 13

That is correct. Can you extend this pattern and my example to what you think the answer should be for the overall problem?

Answer 14

13

Answer 15

Once again, close, but not quite. Recall that \(n!=n\cdot(n-1)\cdot(n-2)\cdot...\cdot3\cdot2\cdot1\). So \[13!=13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\]If we continue the pattern we got earlier, we would have had \[12!=12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\]

So the solution should rather be D. \(12!\)

Answer 16

was i the only one who chuckled after reading the first sentence of the problem?