Hi
Here is the situation:
I mark scissors and look for a system which is both practical and with
many permutations.
They way I want to do it is as following:
At the pivot, i.e. the hole through which the pivot screw goes, at the
inside of one of the blades, I want to do as follows:
The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
each of these 4 positions I have space to make three

small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
- total of 4 markings.
My question is what would be the most efficient way of 'populating'
these areas with either empty spaces or holes and what

is the limit of my permutations around one hole - obviously without
any duplications?
Regards
Gabriel Smit
South Africa

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------

In article ,
 GG  wrote:

> Hi
> Here is the situation:
> I mark scissors and look for a system which is both practical and with
> many permutations.
> They way I want to do it is as following:
> At the pivot, i.e. the hole through which the pivot screw goes, at the
> inside of one of the blades, I want to do as follows:
> The hole has a N, S, W and E or 12 o clock, 3, 6, and 9 o clock. At
> each of these 4 positions I have space to make three
> 
> small holes with a diamond drill - i.e. either nothing or 1 or 2 or 3
> - total of 4 markings.
> My question is what would be the most efficient way of 'populating'
> these areas with either empty spaces or holes and what
> 
> is the limit of my permutations around one hole - obviously without
> any duplications?

You have 4 positions, each of which can have one of 4 markings.  So you 
have 4^4 = 256 possible markings.


> Regards
> Gabriel Smit
> South Africa
-- 
---------------------------
|  BBB                b    \     Barbara at LivingHistory stop co stop uk
|  B  B   aa     rrr  b     |
|  BBB   a  a   r     bbb   |    Quidquid latine dictum sit,
|  B  B  a  a   r     b  b  |    altum viditur.
|  BBB    aa a  r     bbb   |   
-----------------------------