How many distinct ways can you rearrange all the letters of the word BOOKKEEPER?

151,200 distinct arrangements.

BOOKKEEPER has 10 letters, but some repeat: O appears twice, K appears twice, and E appears three times (B, P, and R each appear once). This is a permutation of a multiset, so you can't just use 10!, because swapping two identical letters doesn't produce a new arrangement.

Formula (permutations of a multiset):

count = n! / (k₁! · k₂! · … · kᵣ!)

where n is the total number of letters and each kᵢ is how often a repeated letter occurs.

Calculation:

10! / (2! · 2! · 3!) = 3,628,800 / (2 · 2 · 6) = 3,628,800 / 24 = 151,200

Why divide: 10! counts every ordering as if all letters were distinguishable. Each genuine arrangement gets counted 2! times from the interchangeable O's, 2! times from the K's, and 3! times from the E's — for 24 duplicate listings per distinct word. Dividing by 24 removes the overcount.

Quick check in Python:

from math import factorial
print(factorial(10) // (factorial(2) * factorial(2) * factorial(3)))  # 151200

Sources:

• https://homework.study.com/explanation/how-many-distinct-permutations-can-be-formed-using-the-letters-of-the-word-bookkeeper.html
• https://brainly.com/question/10072873
• https://en.wikipedia.org/wiki/Permutation
• https://www.geeksforgeeks.org/maths/permutations-and-combinations-of-a-multiset/
• https://www.cuemath.com/permutation-formula/