Counting arrangements by splitting into subcases and using complementary counts, including treating items that must stay together as one block and placing items in separate slots to keep them apart.
When counting scenarios that have distinct "sub-cases", we count each case separately, and add the results.
When counting arrangements where certain items need to stay together, we treat those items as one item, and calculate the number of permutations. Then, we multiply by the number of ways the group can be internally ordered.
For example, if we have A, B, C and D but A, B & C must be together, then we have
But since [ABC] can be internally ordered 3=6 ways, we have 12 total arrangements
When counting arrangements where certain items need to stay apart, we first consider the number of ways to arrange the unrestricted items. Then we imagine inserting the remaining letters into different "slots" between the regular items.
For example, if we have A, B, C and D but A, B cannot be adjacent to each other, then we have
of the unrestricted letters. Then A can go in any of the 3 slots, then B in any of the remaining 2 which makes 3×2=6 ways to inert A and B. Thus there are a total of 2×6=12 arrangements.
Imagine a scenario that can be split into two cases:
case A can be done in n ways
case B can be done in k ways
The addition of n+k must give the total count for the scenario. This is very useful for solving counting questions, as it allows us to subtract a number of undesirable cases from the number of total cases to find our desired answer.
Nice work completing Counting with Restrictions, here's a quick recap of what we covered:
Exercises checked off
Counting arrangements by splitting into subcases and using complementary counts, including treating items that must stay together as one block and placing items in separate slots to keep them apart.
When counting scenarios that have distinct "sub-cases", we count each case separately, and add the results.
When counting arrangements where certain items need to stay together, we treat those items as one item, and calculate the number of permutations. Then, we multiply by the number of ways the group can be internally ordered.
For example, if we have A, B, C and D but A, B & C must be together, then we have
But since [ABC] can be internally ordered 3=6 ways, we have 12 total arrangements
When counting arrangements where certain items need to stay apart, we first consider the number of ways to arrange the unrestricted items. Then we imagine inserting the remaining letters into different "slots" between the regular items.
For example, if we have A, B, C and D but A, B cannot be adjacent to each other, then we have
of the unrestricted letters. Then A can go in any of the 3 slots, then B in any of the remaining 2 which makes 3×2=6 ways to inert A and B. Thus there are a total of 2×6=12 arrangements.
Imagine a scenario that can be split into two cases:
case A can be done in n ways
case B can be done in k ways
The addition of n+k must give the total count for the scenario. This is very useful for solving counting questions, as it allows us to subtract a number of undesirable cases from the number of total cases to find our desired answer.
Nice work completing Counting with Restrictions, here's a quick recap of what we covered:
Exercises checked off