Question 1036525: Hi could you please help with this question , i have attempted to try solving it and i got 12 but i am not entirely sure if i did it right
there are three cages arranged in a row. In how many ways can one put a lion, a sheep, and a cow into these cages so that no animal gets eaten?
Thanks
Found 2 solutions by Alan3354, jim_thompson5910:
Answer by Alan3354(69443)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! Let's have three cages A,B,C.
The lion must be in a cage on his own. Otherwise the lion will eat either the sheep or the cow.
Let's say the lion is in cage A. That leaves 2 cages left. If you can put more than one animal to a cage, then cage B could have the sheep and the cow at the same time. In notation form, I'll write this as
{lion}, {sheep,cow},{ }
The empty brackets indicate the cage is empty
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Similarly, we could have this
{lion}, { }, {sheep, cow}
to indicate cage B is empty with cage C with the two other animals.
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Finally, we could have this
{lion}, {sheep}, {cow}
or
{lion}, {cow}, {sheep}
Take note how the curly braces surround each individual animal to indicate they get their own separate cage. The order is important
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The following cases
{lion}, {sheep,cow},{ }
{lion}, { }, {sheep, cow}
{lion}, {sheep}, {cow}
{lion}, {cow}, {sheep}
are where the lion is in cage A. The sheep and the cow are the ones that change on each case basis.
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There are 4 cases where the lion is in cage A. Imagine the same set of cases are laid out again but this time the lion is in cage B (not cage A). The same cases would happen, but we'd have this instead
{sheep,cow}, {lion},{ }
{ }, {lion}, {sheep, cow}
{sheep}, {lion}, {cow}
{cow}, {lion}, {sheep}
So that's four more cases
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Finally, the last set of cases are where the lion is in cage C
{sheep,cow}, { }, {lion}
{ }, {sheep,cow}, {lion}
{sheep}, {cow}, {lion}
{cow}, {sheep}, {lion}
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To neatly sum things up, we have these 12 cases (4*3 = 12)
{lion}, {sheep,cow},{ }
{lion}, { }, {sheep, cow}
{lion}, {sheep}, {cow}
{lion}, {cow}, {sheep}
{sheep,cow}, {lion},{ }
{ }, {lion}, {sheep, cow}
{sheep}, {lion}, {cow}
{cow}, {lion}, {sheep}
{sheep,cow}, { }, {lion}
{ }, {sheep,cow}, {lion}
{sheep}, {cow}, {lion}
{cow}, {sheep}, {lion}
So there are 12 ways to arrange the animals in the cages. This is of course, assuming that the sheep and cow can share a cage. For obvious reasons, the lion gets his own cage. If they cannot share a cage, and there is one animal per cage, then there would be 3*2*1 = 6 possible outcomes.
Side note: if the cage order is important, then the final answer is 12. If the order is not important, then the three subblocks above are pratically the same. So if order isn't important, then there are only 2 cases. Those 2 cases would be
{lion}, {sheep,cow},{ }
{lion}, {sheep}, {cow}
Unfortunately the instructions weren't clear on if order mattered or not.
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