Do three-saw plans still exist on here?

[Derf]

By my math, you get 11 different 3-saw plans....

Actually, there are many more permutations than 11 different 3-saw plans if you have 33 saws to choose from.
1, 2, 3.
2, 3, 4.
3, 4, 5.... etc. That would get you at least 31 3-saw plans...
Then you have
1, 2, 4.
1, 2, 5.
1, 2, 6.... etc.
Then you have
1, 3, 4.
1, 3, 5....
One could say that a permutation is an ordered combination. The number of permutations of n objects taken r at a time is determined by the formula: P ( n , r ) = n ! ( n − r ) ! but if order isn't important and repetition isn't allowed, the formulas would be (n !) / r! (n-r)!

The math works out to 5456 possible 3-saw plans if you have 33 saws. (I was a math minor in a previous life).

 

[PA Dan]

Actually, there are many more permutations than 11 different 3-saw plans if you have 33 saws to choose from.
1, 2, 3.
2, 3, 4.
3, 4, 5.... etc. That would get you at least 31 3-saw plans...
Then you have
1, 2, 4.
1, 2, 5.
1, 2, 6.... etc.
Then you have
1, 3, 4.
1, 3, 5....
One could say that a permutation is an ordered combination. The number of permutations of n objects taken r at a time is determined by the formula: P ( n , r ) = n ! ( n − r ) ! but if order isn't important and repetition isn't allowed, the formulas would be (n !) / r! (n-r)!

The math works out to 5456 possible 3-saw plans if you have 33 saws. (I was a math minor in a previous life).

Have a little time on your hands...?

 

[AlfA01]

Actually, there are many more permutations than 11 different 3-saw plans if you have 33 saws to choose from.
1, 2, 3.
2, 3, 4.
3, 4, 5.... etc. That would get you at least 31 3-saw plans...
Then you have
1, 2, 4.
1, 2, 5.
1, 2, 6.... etc.
Then you have
1, 3, 4.
1, 3, 5....
One could say that a permutation is an ordered combination. The number of permutations of n objects taken r at a time is determined by the formula: P ( n , r ) = n ! ( n − r ) ! but if order isn't important and repetition isn't allowed, the formulas would be (n !) / r! (n-r)!

The math works out to 5456 possible 3-saw plans if you have 33 saws. (I was a math minor in a previous life).

Yeah, but after all that calculating to sort out which three, I'd be to tired to cut wood.

 

[Firewood Bandit]

To answer the original question, ahhhhhhhhh..............NO.

 

[treesmith]

My first thought was 33 x 32 x 31

33 possible first choices, then 32 possible second choices then 31 possible third choices

 

[Derf]

My first thought was 33 x 32 x 31

33 possible first choices, then 32 possible second choices then 31 possible third choices

n! (Pronounced n factorial) is basically that. For example, 3! is 3 x 2 x 1

You're on the right track with that logic, but it isn't quite right. If you had 5 saws, you could choose
1, 2, 3
1, 2, 4
1, 2, 5
1, 3, 4
1, 3, 5
1, 4, 5
2, 3, 4
2, 3, 5
2, 4, 5
Or 3, 4, 5
ten 3-saw plans
You wouldn't get that answer if you did
5 x 4 x 3 x 2 x 1 = 120

But 5! / 3!(5-3)! = 120 / 6*2= 120 / 12 = 10

 

[danimal]

n! (Pronounced n factorial) is basically that. For example, 3! is 3 x 2 x 1

You're on the right track with that logic, but it isn't quite right. If you had 5 saws, you could choose
1, 2, 3
1, 2, 4
1, 2, 5
1, 3, 4
1, 3, 5
1, 4, 5
2, 3, 4
2, 3, 5
2, 4, 5
Or 3, 4, 5
ten 3-saw plans
You wouldn't get that answer if you did
5 x 4 x 3 x 2 x 1 = 120

But 5! / 3!(5-3)! = 120 / 6*2= 120 / 12 = 10

You damn math heads make my brain hurt.
Ain't it simpler to just have a dozen saws and go cut wud?

 

[Derf]

Pretty much

 

[mdavlee]

I’m on a 2.5 saw plan. I have 2 that run and one in a box [emoji23]

 

[00wyk]

I use whichever saw the wood tells me to use.

 

[wcorey]

The math works out to 5456 possible 3-saw plans if you have 33 saws. (I was a math minor in a previous life).

The flaw in that reasoning is that you'd end up with many plans consisting of saws with redundant displacements, not really in keeping with the whole point of a three saw plan...

 

[Sleeper]

What's a three saw plan?

I think at least three saws of each .1 cubic centimeter is needed.