If is a finite set, let be its cardinality. Let be the set consisting of all -digit integers, each digit of which is or . Compute . Let . Compute .
So ? And ?
That does not count all such arrangements though. That counts all 5 digit numbers of the form:
AAABC, where A,B,C are distinct and either 1,2, or 3.
You could also have something of the form ABBCA, which is not counted by your method.
You want to find those 5 digit numbers that have at least one 1, one 2, and one 3.
It is easier to find those that don't have at least one of each, and then subtracting it from
So try to count the following:
# of 5 digit numbers with only 1's.
# of 5 digit numbers with only 2's.
# of 5 digit number with only 3's.
# of 5 digit numbers with only 1's and 2's.
# of 5 digit numbers with only 2's and 3's.
# of 5 digit numbers with only 1's and 3's.
# of 5 digit numbers with only 1's:
# of 5 digit numbers with only 2's:
# of 5 digit number with only 3's:
# of 5 digit numbers with only 1's and 2's:
# of 5 digit numbers with only 2's and 3's:
# of 5 digit numbers with only 1's and 3's:
For the above...would counting the number of those numbers that do not have those numbers and then subtract it be the best way? Because you could have
Suppose we want to count the # of 5 digit numbers with only 1's and 2's:
Such a number can have one 1, two 1's, three 1's, or four 1's. (All others being 2)
How many ways can it have one 1? You have to choose 1 position out of 5 possible positions, so .
How many ways can it have two 1's? You have to choose 2 positions out of 5 possible positions, so .
Similarly for the others, we have 10 and 5. So there are 30.
Do the same for all others, subtract, and you have your result. (150)
If you don't like this approach, you can try counting them directly by splitting them into different cases.
two 1's, two 2's, one 3
two 1's, two 3's, one 2
two 2's, two 3's, one 1
one 1, one 2's, three 3's
one 2, one 3, three 1's
one 1, one 3, three 2's.
Of course, the first three are equal, and the last three are equal.
I'm pretty sure the first way is the simplest way to do it though. I could be wrong.