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wythagoras
wythagoras
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NOTE: Too much for spoiler tags, so I only did the last.

EDIT: After all that work, I find out I have the same answer as wythagoruswythagoras.

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---

Therefore, there are 19 digits with 5 lights.

##Four lights## There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this.

Lastly, you can have two vertical bars disconnected by turning off all three horizontal lights.

Thus, the total number of ways to have four lights is 35-8-2-4-1=20.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

So, the total is:

$$1+7+19+20+16+10+7=80$$

NOTE: Too much for spoiler tags, so I only did the last.

EDIT: After all that work, I find out I have the same answer as wythagorus.

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---

Therefore, there are 19 digits with 5 lights.

##Four lights## There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this.

Lastly, you can have two vertical bars disconnected by turning off all three horizontal lights.

Thus, the total number of ways to have four lights is 35-8-2-4-1=20.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

So, the total is:

$$1+7+19+20+16+10+7=80$$

NOTE: Too much for spoiler tags, so I only did the last.

EDIT: After all that work, I find out I have the same answer as wythagoras.

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---

Therefore, there are 19 digits with 5 lights.

##Four lights## There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this.

Lastly, you can have two vertical bars disconnected by turning off all three horizontal lights.

Thus, the total number of ways to have four lights is 35-8-2-4-1=20.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

So, the total is:

$$1+7+19+20+16+10+7=80$$

added 218 characters in body
Source Link
Trenin
Trenin
  • 9k
  • 22
  • 53

NOTE: Too much for spoiler tags, so I only did the last.

EDIT: After all that work, I find out I have the same answer as wythagorus.

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---

Therefore, there are 19 digits with 5 lights.

##Four lights## With four lights on, three are off. ThereThere are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this.

Lastly, you can have two vertical bars disconnected by turning off all three horizontal lights. 

Thus, the total number of ways to have four lights is 35-8-2-4=214-1=20.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

So, the total is:

$$1+7+19+21+16+10+7=81$$$$1+7+19+20+16+10+7=80$$

NOTE: Too much for spoiler tags, so I only did the last.

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---

Therefore, there are 19 digits with 5 lights.

##Four lights## With four lights on, three are off. There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this. Thus, the total number of ways to have four lights is 35-8-2-4=21.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

So, the total is:

$$1+7+19+21+16+10+7=81$$

NOTE: Too much for spoiler tags, so I only did the last.

EDIT: After all that work, I find out I have the same answer as wythagorus.

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

 ---     ---  
        |   | 
 ---     ---  
|   |         
 ---     ---

Therefore, there are 19 digits with 5 lights.

##Four lights## There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this.

Lastly, you can have two vertical bars disconnected by turning off all three horizontal lights. 

Thus, the total number of ways to have four lights is 35-8-2-4-1=20.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

So, the total is:

$$1+7+19+20+16+10+7=80$$

added 218 characters in body
Source Link
Trenin
Trenin
  • 9k
  • 22
  • 53

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

There##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

Working up##Four lights## With four lights on, therethree are 7 digitsoff. There are $_7C_4=35$ ways to do this. However, with 1three lights off, there are a few ways you can have a disconnected light trivially.

With As before, we saw two ways in which there were disconnected lights, it is a little trickier with five lights. You In each of those, we can have 4 "L" shaped pieces around every corner onturn off a light in the top"o" section and another 4 on the bottomstill be disconnected. This is a total of 8 ways to disconnect. Also, if you can have 2 vertical digits; one on the left and oneturn on the righttwo lights in opposite corners they will be disconnected. Total is 10 There are 2 ways to do this.

All that is left isAlso, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to determine three lights anddo this. Thus, the total number of ways to have four lights is 35-8-2-4=21.

With##Three lights## With three lights, you can have "C"sa "C" and "U"s"U" and "n"s"n" and backwards "C"s"C", both on the top and on the bottom for a total of 84x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

With four lights on, three are off. There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you##Two lights## You can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in4 "L" shaped pieces around every corner on the "o" sectiontop and still be disconnected. This is a total of 8 ways to disconnectanother 4 on the bottom. Also, if you turncan have 2 vertical digits; one on the two lights in opposite corners they will be disconnected. There are 2 ways to do this. Thus,left and one on the total number of ways to have four lightsright. Total is 2510.

##One light## There are 7 digits with 1 light trivially.

$$1+7+19+25+16+10+7=85$$$$1+7+19+21+16+10+7=81$$

If all seven lights are on, there is one possible digit.

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

Working up, there are 7 digits with 1 light trivially.

With two lights, it is a little trickier. You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

All that is left is to determine three lights and four lights.

With three lights, you can have "C"s and "U"s and "n"s and backwards "C"s, both on the top and on the bottom for a total of 8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right, or its reverse, for a total of 2 more. Grand total is 16.

With four lights on, three are off. There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this. Thus, the total number of ways to have four lights is 25.

$$1+7+19+25+16+10+7=85$$

##Seven lights##

If all seven lights are on, there is one possible digit.

##Six lights##

If six lights are on, there are 7 ways you can remove a single light from the "all on" case, so there are 7 digits with six lights.

##Five lights## There are $_7C_5=21$ ways to get digits with 5 lights, but a couple of them (2) are disconnected.

##Four lights## With four lights on, three are off. There are $_7C_4=35$ ways to do this. However, with three lights off, there are a few ways you can have a disconnected light. As before, we saw two ways in which there were disconnected lights with five lights. In each of those, we can turn off a light in the "o" section and still be disconnected. This is a total of 8 ways to disconnect. Also, if you turn on the two lights in opposite corners they will be disconnected. There are 2 ways to do this.

Also, you can orphan a vertical piece as follows;

 ---         ---
|   |       |
 ---    -->  
|   |       |   |
 ---

There are 4 ways to do this. Thus, the total number of ways to have four lights is 35-8-2-4=21.

##Three lights## With three lights, you can have a "C" and "U" and "n" and backwards "C", both on the top and on the bottom for a total of 4x2=8. Also, you can have "7" and "L" and their reverse for a total of 4. You can have a left "T" and a right "T" for 2 more. And finally, you can have an "H" missing the top left and bottom right (kind of a zig zag), or its reverse, for a total of 2 more. Grand total is 16.

##Two lights## You can have 4 "L" shaped pieces around every corner on the top and another 4 on the bottom. Also, you can have 2 vertical digits; one on the left and one on the right. Total is 10.

##One light## There are 7 digits with 1 light trivially.

$$1+7+19+21+16+10+7=81$$

Source Link
Trenin
Trenin
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