How many distinct arrangements are possible using the letters of (PROBABILITY)?

Correct answer: (9979200)

The direct answer is option A: 9,979,200 distinct arrangements. The word has 11 letters. The letters B occur twice and I occur twice; the other seven letters occur once each. If all 11 positions were treated as different, there would be \(11!\) arrangements. However, swapping the two B's does not create a new visible arrangement, so divide by \(2!\); similarly divide by \(2!\) for the two I's. Therefore the answer is \(\frac{11!}{2!2!}=\frac{39,916,800}{4}=9,979,200\). Option A is correct. Option B is \(11!\), so it incorrectly treats repeated letters as distinct. Option C divides by too much or otherwise misses valid arrangements. Option D is exactly \(11!\), again ignoring repetitions. The important rule is: for repeated objects, divide by the factorial of each repetition count. Memory cue: distinct arrangements equal total factorial divided by repeated-letter factorials.