Competition problem — Combinatorics

Combinatorics Difficulty 2.8 Junior Find the answer Philippines

Problem:

In how many ways can the letters of the word PANACEA be arranged so that the three As are not all together?
(a) 540
(b) 576
(c) 600
(d) 720

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is d, and the solution below works it through.

Solution

Solution:

The word PANACEA has 7 letters, with the letter A appearing 3 times, and the other letters P, N, C, E each appearing once.

First, find the total number of arrangements of the letters:

Number of arrangements $= \dfrac{7!}{3!} = \dfrac{5040}{6} = 840$

Now, count the number of arrangements where all three As are together.

Treat the three As as a single letter (block), so we have: [AAA], P, N, C, E — a total of 5 objects to arrange.

Number of arrangements $= 5! = 120$

Therefore, the number of arrangements where the three As are NOT all together is:

$840 - 120 = 720$

So, the answer is (d) 720.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.