Difference between revisions of "2024 SSMO Speed Round Problems/Problem 4"
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==Problem== | ==Problem== | ||
| − | Sam wants to read the | + | Sam wants to read the <i>Harry Potter</i> and <i>Warriors</i> books. There are 7 <i>Harry Potter</i> books that must be read in a specific order, and there are 6 <i>Warriors</i> books that also must be read in a specific order; however, he can read the two series at the same time. For example, he could read the first three <i>Harry Potter</i> books, then the first five <i>Warriors</i> books, then the remaining <i>Harry Potter</i> books, and finally the last <i>Warriors</i> book. In how many unique orders can Sam read the books? |
==Solution== | ==Solution== | ||
| + | Note that there are <math>13!</math> ways to order the books. However, out of the <math>7!</math> possible ways to order the 7 Harry Potter books we included, only one actually works. Similarly, out of the <math>6!</math> possible ways to order the <math>6</math> Warrior books we included, only one actually works. So, the answer is <cmath>\frac{13!}{7!6!} = \frac{13\cdot12\cdot11\cdot10\cdot9\cdot8}{6!=720} = 13\cdot12\cdot11 = \boxed{1716}.</cmath> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:26, 10 September 2025
Problem
Sam wants to read the Harry Potter and Warriors books. There are 7 Harry Potter books that must be read in a specific order, and there are 6 Warriors books that also must be read in a specific order; however, he can read the two series at the same time. For example, he could read the first three Harry Potter books, then the first five Warriors books, then the remaining Harry Potter books, and finally the last Warriors book. In how many unique orders can Sam read the books?
Solution
Note that there are 
ways to order the books. However, out of the 
possible ways to order the 7 Harry Potter books we included, only one actually works. Similarly, out of the 
possible ways to order the 
Warrior books we included, only one actually works. So, the answer is ![\[\frac{13!}{7!6!} = \frac{13\cdot12\cdot11\cdot10\cdot9\cdot8}{6!=720} = 13\cdot12\cdot11 = \boxed{1716}.\]](http://latex.artofproblemsolving.com/2/4/f/24f22bbd3a768c554766cb844d21238e7a8c7045.png)
~SMO_Team
