Any Mathematicians?

[jimcross]

Any mathematicians here? With 10 pins, how many different sequences, of pins falling, can result in a strike?

For instance, in a " perfect " strike, the ball hits 4 pins,- 1, 3, 5 and 9.
The 1 takes out the 2, 4, and 7. The 3 takes out the 6 and 10, and the 5 takes out the 8.

I once saw a little kid put the ball down s-l-o-w-l-y, and you could have gone and bought a drink while it was going down the lane. It just grazed the 10 pin as it fell into the gutter. The 10 wobbled a bit, then fell over and just touched the 9. Then, one by one, it continued to happen, one at a time,'til finally something - I don't remember which, fell forward and clipped the headpin. Took a long time, probably couldn't happen on to-day's machines which are camera activated. That one, the ball had to hit the back cushion to start the cycle. It didn't hit it hard enough, so the machine didn't cycle.

1000 combinations, 10,000?

On a similar note - how many ways are there to miss a strike?? Joking ... but I've had most of them!

 

[Hooligan]

Jim, it roused my interest. lol
But be more specific. Can I hit the 8 pin first? The 9 pin, the 5 pin???? Do you just mean the order they fall? If so 10x9x8x7...etc will do it. (3628800)
Or - can a single pin take out 2 pins (or more), and you count that differently?

The answer is @#$@#%$% heaps. lol

As for your joke question - how many ways are there to miss a strike??
Only one... get less than 10 first ball haha.

 

[jason_doust]

Ooh, that'd be 10 factorial! Or if you only count the front 7 pins you can hit, it's 7 factorial, expressed as "7!".

That'd be 5040, btw.

 

[jimcross]

Well Hooligan, that's a hard question to handle, but I've just had a bit of a measure of a bowling ball, and it appears to me, that unless you've got one a lot narrower than any of mine, the only pins you can hit to begin the sequence, will be 1, 2, 3, 4, 6, 7, or 10.

Second: any pin can do anything, ( and frequently does ) once having been hit. Every different sequence counts as one.

Your answer must be wrong, 'cause I don't understand it. Is it algebra?

Finally, if getting less than 10 pins with the first ball is not a strike, do you have to get more than 10 pins with the first ball to get one? No wonder I have trouble.

 

[Hooligan]

Ooh, that'd be 10 factorial! Or if you only count the front 7 pins you can hit, it's 7 factorial, expressed as "7!".

That'd be 5040, btw.

Jason, as you can only start the sequence with 7 of the pins, yet there are 10 pins in total, would the answer not be

7 x 9! = 2540160

?

 

[nls]

I would think it would be 10 factorial less 3 factorial (god knows what that equals....). I don't think 9 factorial x 7 would give you the same answer.

You need to work out the total number of possibilities (10 factorial) and then subtract the number of possibilities that start with the 5, 8 & 9 pins (ie 3 factorial)....

 

[TheMatrix]

You guys forgot to consider the ball as well ... Just too many ways to get all 10 pins to fall but it can be done if you really have to

 

[Shayne Melton]

I would think it would be 10 factorial less 3 factorial (god knows what that equals....). I don't think 9 factorial x 7 would give you the same answer.

You need to work out the total number of possibilities (10 factorial) and then subtract the number of possibilities that start with the 5, 8 & 9 pins (ie 3 factorial)....

What if the ball was lofted to the pin deck and landed on one of those 3 pins on the full? If working with fantasy, all situations must be considered.

 

[jimcross]

Couldn't get past the masking units, high enough. Got to be the total number less the ones starting with 5, 8, or 9.

 

[Shayne Melton]

Couldn't get past the masking units, high enough. Got to be the total number less the ones starting with 5, 8, or 9.

What if I was bowling at a country centre that hasn't got automatic pin setters and still uses pin boys to set the decks?

 

[TheMatrix]

Couldn't get past the masking units, high enough. Got to be the total number less the ones starting with 5, 8, or 9.

Has to be more .. because you assume the pins will take the other pins down. What if they don't and the pins are knocked off by the ball !

 

[Roysa]

ive got a headache...

 

[Hooligan]

What if one of the pins stood back up, then was knocked over again by a pin other than the one that knocked it over first time, then it knocked over some combination of pins?
Roysa, how's the headache now?