(related to Problem: The Three Sheep)
The number of different ways in which the three sheep may be placed so that every pen shall always be either occupied or in line with at least one sheep is forty-seven.
The following table, if used with the key in Diagram $1,$ will enable the reader to place them in all these ways:—
|Two Sheep||Third Sheep||No. of Ways|
|A and B||C, E, G, K, L, N, or P||7|
|A and C||I, J, K, or O||4|
|A and D||M, N, or J||3|
|A and F||J, K, L, or P||4|
|A and G||H, J, K, N, O, or P||6|
|A and H||K, L, N, or O||4|
|A and O||K or L||2|
|B and C||N||1|
|B and E||F, H, K, or L||4|
|B and F||G, J, N, or O||4|
|B and G||K, L, or N||3|
|B and H||J or N||2|
|B and J||K or L||2|
|F and G||J||1|
This, of course, means that if you place sheep in the pens marked $A$ and $B,$ then there are seven different pens in which you may place the third sheep, giving seven different solutions. It was understood that reversals and reflections do not count as different.
If one pen at least is to be not in line with a sheep, there would be thirty solutions to that problem. If we counted all the reversals and reflections of these $47$ and $30$ cases respectively as different, their total would be $560,$ which is the number of different ways in which the sheep may be placed in three pens without any conditions. I will remark that there are three ways in which two sheep may be placed so that every pen is occupied or in line, as in Diagrams $2,$ $3,$ and $4,$ but in every case each sheep is in line with its companion. There are only two ways in which three sheep may be so placed that every pen shall be occupied or in line, but no sheep in line with another. These I show in Diagrams $5$ and $6.$ Finally, there is only one way in which three sheep may be placed so that at least one pen shall not be in line with a sheep and yet no sheep in line with another. Place the sheep in $C, E, L.$ This is practically all there is to be said on this pleasant pastoral subject.
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