What to do?

• Oct 23rd 2006, 04:44 PM
Dragon
What to do?
After rolling the first ball of a frame in a game 10-pin bollowing, how many pin configurations are physically possible?
• Oct 23rd 2006, 05:31 PM
ThePerfectHacker
Quote:

Originally Posted by Dragon
After rolling the first ball of a frame in a game 10-pin bollowing, how many pin configurations are physically possible?

Consider the following diagram,
Code:

  *   **  *** ****
In each star there are 10, but after you use 1 pin there are 9 left. For the next star there are 9 choices and so on....
Thus,
$\displaystyle 10\cdot 9\cdot 8\cdot ... \cdot 2\cdot 1=10!$
• Oct 24th 2006, 02:45 PM
Rimas
what?
• Oct 28th 2006, 11:13 AM
ceasar_19134
Me and dragon are friends, and he didnt post the correct problem.

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)?

Assuming ThePerfectHacker's solution is still correct, could somebody explain it a little more clearly?

(I'd make a new thread but I'm not sure if it violates the identical thread rule.)
• Oct 28th 2006, 11:56 AM
Quick
Quote:

Originally Posted by ceasar_19134
Me and dragon are friends, and he didnt post the correct problem.

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)?

Assuming ThePerfectHacker's solution is still correct, could somebody explain it a little more clearly?

(I'd make a new thread but I'm not sure if it violates the identical thread rule.)

Try this:

Draw the diagram of the pins:
Code:

o o o o  o o o   o o   o
Now knock over a pin:
Code:

o o o o  o o o   x o   o
That's 1 combination, now put that pin back up and knock over another pin:
Code:

o o x o  o o o   o o   o
That's another combination, since you can knock over ten individual pins there are 10 physical combinations of pins when 1 is knocked over...

Now what if 2 were knocked over?

Well first thing to do is to draw the the pins with 1 knocked over:
Code:

o o o o  o o o   o o   x
Now, knock another one over:
Code:

o o o o  o o o   x o   x
Put it back up and knock another one over:
Code:

o o o o  o o x   o o   x
If you keep doing this you'll realize that there are nine combinations..

Thus you have 9 combinations per pin, since you have 10 pins you have 9(10) combinations for the whole thing, which is 90

However every one of those combinations is counted twice, notice that if I knock down a pin:
Code:

o o o o  o x o   o o   o
and knocked down another pin:
Code:

o o o o  o x o   x o   o
it is the same combination as if I knocked down this pin:
Code:

o o o o  o o o   x o   o
and then knocked down the second pin:
Code:

o o o o  o x o   x o   o
So the number of combinations for 2 pins is 90/2, which equals 45

So far you have 10+45=55 combinations

more to come (going to do it in another post)...
• Oct 28th 2006, 12:02 PM
Quick
Quote:

Originally Posted by Quick
So the number of combinations for 2 pins is 90/2, which equals 45

Now onto 3 pins, knock over 2 pins:
Code:

o o o o  o o o   x o   x
Now knock over another one:
Code:

o o o x  o o o   x o   x
Reset and do it again:
Code:

o x o o  o o o   x o   x
You might realize that there are 8 possible combinations.

Thus the number of combinations for three pins is the number of combinations for 2 pins times 8, which is 45*8=360

But wait! each combination is counted three times! so you divide the number by 3, thus you have 360/3=120 combinations

Add that to the amount of combinations you have so far to get: 10+45+120=175

Now you try doing the amount of combinations for 4, 5, 6, 7, 8, 9, and 10 pins, and solve from there!
• Oct 28th 2006, 12:44 PM
Quick
Quote:

Originally Posted by ceasar_19134
Assuming ThePerfectHacker's solution is still correct, could somebody explain it a little more clearly?

(I'd make a new thread but I'm not sure if it violates the identical thread rule.)

1) It would have violated the identical thread rule

2) I don't think PH's solution is correct