Permutations with Repeats

<p><a href="https://www.prepswift.com/quizzes/quiz/prepswift-permutations-with-repeats" target="_blank">Permutations with Repeats Exercise</a></p><p>How do we deal with <strong><span style="color:#8e44ad;">Permutations with Repeats</span></strong>?</p> <p><strong><span style="color:#e74c3c;">Example</span></strong></p> <p><i>In how many distinct ways can the letters of the word EGG be arranged?</i></p> <p style="margin-left: 40px;"><strong><u>Expectation</u></strong>: Simply do the following $3 \times 2 \times 1 = 6$.</p> <p>But this doesn&#39;t work. There are only THREE distinct arrangements: EGG, GEG, and GGE. Why? Because of that damn repeated G.</p> <p><strong><span style="color:#2980b9;">The Formula</span></strong></p> <p>$$\frac{n!}{(repeat_1)!(repeat_2)!(repeat_3)!}$$</p> <p>This formula accounts for repeats. In our original example, notice that the G was repeated twice, so the formula becomes:</p> <p>$$\frac{3!}{2!} = 3$$</p> <p><strong><span style="color:#e74c3c;">Example 2</span></strong></p> <p>How about in the case of GOOGLE? Notice that there are two Gs and two Os, so the formula is:</p> <p>$$\frac{6!}{2!2!} = 180$$</p> <p><strong><span style="color:#e74c3c;">Example 3</span></strong></p> <p>What about something silly like AAABBBCC? Notice that there are three As, three Bs, and two Cs, so the formula becomes:</p> <p>$$\frac{8!}{3!3!2!}=560$$</p>