We wish to create a passcode that must contain 3 A's, 3 B's, and 3 C's (for example, ABACCBABC would be one of these). How many different passcodes (order matters) could we create? HINT: This is a permutation of like objects.1680100051227
Question
We wish to create a passcode that must contain 3 A's, 3 B's, and 3 C's (for example, ABACCBABC would be one of these). How many different passcodes (order matters) could we create? HINT: This is a permutation of like objects.1680100051227
Solution
The problem is asking for the number of permutations of a multiset. This is a common problem in combinatorics, a branch of mathematics concerned with counting and arrangement.
The formula for the number of permutations of a multiset is:
n! / (r1! * r2! * ... * rk!)
where n is the total number of items, and r1, r2, ..., rk are the counts of each type of item.
In this case, we have 9 items total (n=9), and 3 each of A, B, and C (r1=r2=r3=3).
So the number of permutations is:
9! / (3! * 3! * 3!) = 1680
So there are 1680 different passcodes that can be created.
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