math extra credit?

Question

After rolling the first ball of a frame in a game of 10-pin bowling, how many different pin configurations can remain (assuming all configurations are physically possible)?

Answer 1

Each pin could be "up" or "down". That would be 10 pins, with two choices for each pin. That would multiply to 2^10, or 1024 possibilities.

Answer 2

I am not 100% on this, but it seems like it should be something like
if there is
1 left: 10 possible places
2 left: 10 x 9 possible places ( 10 possible for first pin, 9 remaining to choose for 2nd pin)
3 left: 10 x 9 x 8 possible places
4 left: 10 x 9 x 8 x 7
5 left: 10 x 9 x 8 x 7 x 6
6 left: 10 x 9 x 8 x 7 x 6 x 5
7 left: 10 x 9 x 8 x 7 x 6 x 5 x 4
8 left: 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3
9 left: 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2
10 left: 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

I dont have a calculator, but i think that can be simplified to
10!( 1/9! + 1/8! + 1/7! + 1/6! + 1/5! + 1/4! + 1/3! + 1/2! + 1/1!)

Once again, just a guess, assuming that the pins can be potentially completely mixed after bowling the ball -- and this represents the number of completely unique positions that they could all swap to, even though i imagine it is physically impossible to not knock down a pin and get them to swap spots.

Answer 3

Sounds like a binary problem:
Think of a pin up as 0 and pin down as 1

Then the combinations would be as follows:

0000000000
0000000001
...
1111111111

The total number of combinations would be 2^10 = 1024

Look at an example with 3 pins:

000
001
010
011
100
101
110
111

That would be 8 combinations or 2^3 =D

Answer 4

The pin can either be up, or it can be down. This is a binary operation. You figure these out by saying 2 to the what power? The what power is how many objects are binary. 10 pins gives 2^10 = 1024

Answer 5

2^10 = 1024, because that is the number of ways to choose k pins out of the 10 to remain standing, for k=0,1,...,10.

Answer 6

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Answer 7

2^10 or 1024.