high School⏱ 4 min

Permutations with Repetition

n!/(k1! × k2!)

○Not Started

Arrangements of the multiset {A,A,B} - identical letters are indistinguishable:

2× A

1× B

→ 3

All 3 distinct arrangements:

AAB

ABA

BAA

P'(2,1) = \frac{3!}{2! \cdot 1!} = 3

count of A

count of B

count of C

Out of 3! orderings many look identical because repeated letters can't be told apart. Each group of swaps counts once - hence we divide by 2!·1!.