Is This Math Correct?

Johnnyt

Johnnyt

Burn all jump cues
Silver Member
In 10-ball you need to have the 1 in front, the 10 in the mid, and the 2 and 3 on the back corners. That leaves 6 balls that can be placed in the other parts of the rack. How many combinations can those 6 balls be placed in. I came up with 720...that can't be right. Little math help here. Johnnyt
 
Johnnyt said:
In 10-ball you need to have the 1 in front, the 10 in the mid, and the 2 and 3 on the back corners. That leaves 6 balls that can be placed in the other parts of the rack. How many combinations can those 6 balls be placed in. I came up with 720...that can't be right. Little math help here. Johnnyt
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That sounds about right. I think the math thing is 6!, so 6*5*4*3*2*1, which is 720. Haven't done that stuff in years, so I may be wrong..
 
If you're interested in the total number of configurations, you have to multiply by two for the 2/3 left/right choice.
 
Johnnyt said:
In 10-ball you need to have the 1 in front, the 10 in the mid, and the 2 and 3 on the back corners. That leaves 6 balls that can be placed in the other parts of the rack. How many combinations can those 6 balls be placed in. I came up with 720...that can't be right. Little math help here. Johnnyt
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When Orcullo racks , the combination is ONE. He racks the same every time and that it was I saw the other night. 9,8 up front, 7,6 middle, 5,4 back row.
 
Johnnyt said:
In 10-ball you need to have the 1 in front, the 10 in the mid, and the 2 and 3 on the back corners. That leaves 6 balls that can be placed in the other parts of the rack. How many combinations can those 6 balls be placed in. I came up with 720...that can't be right. Little math help here. Johnnyt
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Why do you "need" the 2 & 3 on the back corners?

pj
chgo
 
Patrick Johnson said:
Why do you "need" the 2 & 3 on the back corners?

pj
chgo
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The 2-3 in the corners is not in the rules, not even in the current BCAPL rules that the US Open (which required the 2-3 in the corners) was played by. However, this requirement will be included in the new BCAPL rules effective 6/1. Don't know of any change pending in the WPA rules.
 
Johnnyt said:
In 10-ball you need to have the 1 in front, the 10 in the mid, and the 2 and 3 on the back corners. That leaves 6 balls that can be placed in the other parts of the rack. How many combinations can those 6 balls be placed in. I came up with 720...that can't be right. Little math help here. Johnnyt
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720 is correct. If the 1, 2, 3, and 10 are in the same position every time, and the other 6 balls are in random order, then here's the math.

6! (six factorial) = 6 X 5 X 4 X 3 X 2 X 1 = 720

I see that I'm repeating what someone else already said after I finished typing this, but here it is anyway.

However, I don't know of any 10 ball rules where the 2 and the 3 have to go on the corners. That would give you 8 balls in random positions instead of 6. You'd have 8! or 40,320 possible number combinations.
 
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You guys are scaring me .... To even have that thought enter you mind is amazing, given the equation that follows the break. Which is how many ways the balls can lie on the table.

If my math is right its twelve bazillion squared multiplied by 43 gazillion, plus one to the pukeillion cubed.

Kinda makes the first variable meaningless ... no? :scratchhead:

I still say the best way to start off one of these rotation games is by hanging a chug-a-luck cage over the table and fill it with the balls, three turns of the cage, and let em fly. :D
 

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DogsPlayingPool said:
The 2-3 in the corners is not in the rules, not even in the current BCAPL rules that the US Open (which required the 2-3 in the corners) was played by. However, this requirement will be included in the new BCAPL rules effective 6/1. Don't know of any change pending in the WPA rules.
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Helps prevent pattern racking especially when using the magic rack.
 
Bob Jewett said:
If you're interested in the total number of configurations, you have to multiply by two for the 2/3 left/right choice.
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The position of the 2 and 3 are irrelevant. The same 2 spots are unavailable to the remaining 6 balls. The total number of combinations are 720. However, I don't know of any rules stating the 2 and 3 are in the rear, so the number of combinations is 8!. But to be totally honest, there is really only 360 combinations. The 6 balls in question are in a perfect hexagonal shape. If you split it in half down the middle, you have a mirror image on each side of the line. You have 360 possible combinations on each side.
 
Banks said:
That sounds about right. I think the math thing is 6!, so 6*5*4*3*2*1, which is 720. Haven't done that stuff in years, so I may be wrong..
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Hey Banks just wanted to say thanks I never knew that was the way to do that. I like that Thanks!
 
subdude1974 said:
the position of the 2 and 3 are irrelevant. The same 2 spots are unavailable to the remaining 6 balls. The total number of combinations are 720. However, i don't know of any rules stating the 2 and 3 are in the rear, so the number of combinations is 8!. But to be totally honest, there is really only 360 combinations. The 6 balls in question are in a perfect hexagonal shape. If you split it in half down the middle, you have a mirror image on each side of the line. You have 360 possible combinations on each side.
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lol.....lol

720 is right. I use that to figure superfectas at the horse track all the time
 
this is one of those things that we thought we'd never have a use for when they tried to teach it to us in high school:D:cool:
 
subdude1974 said:
The position of the 2 and 3 are irrelevant. The same 2 spots are unavailable to the remaining 6 balls. The total number of combinations are 720. However, I don't know of any rules stating the 2 and 3 are in the rear, so the number of combinations is 8!. But to be totally honest, there is really only 360 combinations. The 6 balls in question are in a perfect hexagonal shape. If you split it in half down the middle, you have a mirror image on each side of the line. You have 360 possible combinations on each side.
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Well, no. If you're going to discard positions for mirror symmetry, you cannot do it twice. The placement of the 2/3 determines the "sidedness" of the arrangement. After that there are 720 permutations of the remaining 6 balls.
 
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