Math Talent Quest
May 23rd: Casework
Casework is extremely important for solving combinatorics problems. Everyone knows how to do
casework to a certain degree, but in harder math competitions it is crucial to do casework in an
organized fashion. Even more importantly, how one chooses the “cases” directly determines if it
takes 1 hour or 10 minutes to find the answer. In general, less cases is better because more cases
means easier to mess up numbers during the process.
There is no specific formulas for doing “caseworks.” Besides learning basic examples of organized
caseworks, it really depends on how much one practices. The most important detail one should
keep in mind when practicing “caseworks” is keeping the cases very organized. A procedure for
caseworking is listed below:
1. Before you start caseworking, make sure you write out all the possible cases in a list and
understand what you will be facing in each of these cases. Never start caseworking without a
plan.
2. On your paper, make a column or row for each separate case and clearly label what case you
are doing for each column or row.
3. Make sure you check again that you are not missing an “extreme case” (such as when n=0).
4. For harder and more complicated problems, it is always a good idea to check every single
casework after you have finished the problems; casework is one of the easiest method to
make “silly mistakes” such as computing errors.
These rules, of course, are not strict. You might need to elaborate the steps for Olympiad-level
problems and to reduce for easy AMC 10 problems. Sometimes there are more than one way to
do casework, and if they require about the same work, it (obviously) doesn’t matter which way
you choose.
Most important reminders when doing casework are listed below:
1. The caseworks must cover all possible situations.
2. Any two cases must be separate from each other; there can be no overlap.
3. Every time when you separate into “cases,” you must use the same criterion.
4. How you choose the “cases” depend on how much easier are your “cases” than solving the
original problem with brute force.
Example 1:
Using digits 1, 2, 3, 4, and 5 only, how many integers greater than 21300 can we make if each
digit can be used only once.
Solution:
First, obviously the integer we create has have 5 digits.
If we just want a 5-digit integer not necessarily greater than 21300 using these digits, we can
simply do 5!. Now, if the ten-thousand digit is 3, 4, or 5, then the integer we create automatically
satisfies the “21300” restriction. Thus, let’s divide our case into 2 first: those that have a
ten-thousand digit of 3, 4, or 5 and those that have a ten-thousand digit of 2. We are analyzing
the ten-thousand digit here.
1. If the ten-thousand digit is 3, 4, or 5, we can make 4! 4-digits integers from the rest of
the digits. So we make (3)(4!)=72 from this case.
2. If the ten-thousand digit is 2. We need to divide into more cases because the following
4-digits integer we create must be greater than 1300. Obviously we can divide this into 2
cases just like we did before. We analyze the thousands digit this time.
a) If the thousands digit is 3, 4, or 5 (we already used 2), then we have 3! choices for
the rest 3 digits. SO we make a total of (3)(3!)=18 from this case.
b) If the thousands digit is 1, we need to divide into further cases (you can probably
see why). We analyze the hundred digit this time. Obviously the hundreds digit
can’t be 1 or 2.
i.
If the hundreds digit is 4 or 5, then we can just do 2! for the 2 digits left. So
we have (2)(2!)=4 for this case.
ii.
If the hundreds digit is 3, we can still do 2! for the 2 digits left because we
can’t create “00” from any of the 5 digits we have. So we have (1)(2!)=2 for
this case.
Thus, adding all the cases together, we have 72+18+4+2=96 in total, and that’s our answer.
In your solution to this problem, the process would probably only take a couple lines. This is
considered as an easy problem and thus you shouldn’t spend too much time on organizing (as
you don’t want to spend more than 2 minutes on an AMC 12 #12 or so), but you definitely should
form a basic outline of your solution in your head no matter how easy the problem is.
Example 2:
A 2 × 3 rectangular garden is divided into 6 1 × 1 square regions: A, B, C, D, E, F. In each of
these regions we plant a fruit, and no two adjacent squares (diagonals do not count) can have the
same fruit. We have 6 fruits to choose from, how many ways are there to plant the garden?
Solution:
As the figure shows:
A
C
E
B
D
F
This is an example of a typical casework problem that requires thinking before action. A bad case
work strategy can cause extremely complicated work.
1. Let’s try to put fruits into A and B first. Notice that since the garden is still empty, the way we
put the fruits into A and B is not influenced by any other factor. Thus, there are (6)(5)=30
ways to put fruits into A and B.
So boxes A and B have fruits in them now, and all other boxes are still empty.
Now consider C and D.
A. If C chooses the same fruit as B, then D only has 5 choices. That makes (1)(5)
choices for boxes C and D in this case.
B. If C chooses a different fruit from B, then D has 4 choices. That makes (4)(4) choices
for C and D in this case. (Notice that the first “4” comes from the fact that C can’t be
the same as A and C can’t be the same as B.)
So we have a total of (1)(5)+(4)(4)=21 ways to choose for C and D.
Finally, let’s consider E and F.
Since E and F are in the same situation as we chose for “C and D,” we also have 21
choices to plant E and F.
Notice that when you go down a chain of actions (like in this example, we first plant A and B, then
C and D, and then E and F) with one action following another, we multiply the numbers we get
from the corresponding cases (in this problem, the corresponding cases are “A and B,” “C and D,”
“E and F.”) Thus we have (30)(21)(21) choices (the bolded numbers above) overall for the garden.
Example 3:
4 couples go watch a movie. All 8 people sit in one row. If a girl’s neighbor can only be either her
boyfriend or another girl, we many ways can we arrange the 8 people in one row?
Solution:
Let’s proceed by making some obvious observations:
There are 4! ways just to arrange the girls. There must be either 0 or at least 2 boys between any
two girls. Similarly, there must be either 0 or at least 2 girls between any two boys.
We see the continuous girls (with no boy in between any of them) as one group. Then, the ways
to arrange the girls into groups (where order doesn’t matter) are: 4, 3+1, 2+2, 2+1+1, 1+1+1+1.
(Notice how I’m writing down all the possible cases before I start).
Going in order:
1. When 4 girls are sitting together, we have the following options: (G=Girl and B=Boy).
a) GGGGBBBB or BBBBGGGG
b) BBBGGGGB or BGGGGBBB
c) BBGGGGBB
In this case there are 3!, 2!, and 2! ways (respectively to a, b, c above) for the 4 boys to
sit in the “B spots.” Thus, in total there are (2)(3!)+(2)(2!)+2!=18 ways to sit the 4 boys
to the B spots under this case.
2. In the 3+1 case, we have the following options:
a) GGGBBBBG or GBBBBGGG
b) BGGGBBBG or GBBBGGGB
c) BBGGGBBG or GBBGGGBB
In this case there are 2!, 1, 1 ways (respectively to a, b, c) for the 4 boys to sit in the “B”
spots.
Thus in total there are (2)(2!)+(2)(1)+(2)(1)=8 ways to sit the 4 boys in the “B spots” in
the 3+1 case.
3. In the 2+2 case:
a) GGBBBBGG
b) BGGBBBGG or GGBBBGGB
c) BBGGBBGG or GGBBGGBB
d)
BGGBBGGB
In this case there are 2!, 1, 1, 1 ways for boys to sit, in order or a, b, c, d. So in total
there are (1)(2!)+(2)(1)+(2)(1)+(1)=7 ways to sit the 4 boys into the “B” spots under 2+2
case.
4. In the 2+1+1 case
a) GBBGGBBG
In this case there can only be one way for boys to sit.
5. In the 1+1+1+1 case, it’s impossible to form any possible arrangement because we would
need at least (2)(3)=6 boys.
Because the way to arrange the 4 girls is 4!=24 (as we figured in the beginning), the total way to
sit the boys and girls is:
24×(18+8+7+1)=816.
This is considered as a moderately complicated problem. It is very important to have a very
organized idea of the “caseworks” before you head into “caseworking.” During caseworking, it is
still important to keep everything organized so you don’t miss any sub-cases and so it’s easier for
you to check your work later. I can’t emphasize how important it is to keep the cases organized!