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Counting with Restrictions
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Advanced techniques for scenarios with branches, negative counting, and permutations where items need to stay together / separate.
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Counting with subcases
AHL AA 1.10
When counting scenarios that have distinct "sub-cases", we count each case separately, and add the results.
Orderings where certain items need to stay together
AHL AA 1.10
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
[ABC]D or D[ABC]→2 arrangements
But since [ABC] can be internally ordered 3=6 ways, we have 12 total arrangements
Orderings where certain items need to stay apart
AHL AA 1.10
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
XCXDX,XDXCX→2 arrangements
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.
Complementary counts
AHL AA 1.10
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.
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