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- Oct 23rd 2006, 04:44 PM #1

- Oct 23rd 2006, 05:31 PM #2

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- Oct 24th 2006, 02:45 PM #3

- Oct 28th 2006, 11:13 AM #4

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Me and dragon are friends, and he didnt post the correct problem.

It should read:

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 #5
Try this:

Draw the diagram of the pins:

Code:o o o o o o o o o o

Code:o o o o o o o x o o

Code:o o x o o o o o o o

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

Code:o o o o o o o x o x

Code:o o o o o o x o o x

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

Code:o o o o o x o x o o

Code:o o o o o o o x o o

Code:o o o o o x o x o o

So far you have 10+45=55 combinations

more to come (going to do it in another post)...

- Oct 28th 2006, 12:02 PM #6
Now onto 3 pins, knock over 2 pins:

Code:o o o o o o o x o x

Code:o o o x o o o x o x

Code:o x o o o o o x o x

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 #7