[Math] How many ways can $2$ different history books, $5$ different math books, and $4$ different novels be arranged on a shelf if …

combinatoricspermutations

This is my first class in probability so I just wanted verification as to my attempted solution.

Question:

In how many ways can $2$ different history books, $5$ different math books, and $4$ different novels be arranged on a shelf if the books of each type must be together?

Solution:

$2! \cdot 5! \cdot 4! \cdot 3! = 34560$

There are $34560$ ways the books can be arranged.

Best Answer

Question: "In how many ways can 2 different history books, 5 different math books, and 4 different novels be arranged on a shelf if the books of each type must be together?"

In this question, sequence of the books is not important, therefore:

  • For the 2 history books: 2 ways to arrange them (AB and BA), or $2!$
  • For the 5 math books: $5*4*3*2*1 = 5!$ ways to arrange them, or 120
  • For the 4 novels: $4*3*2*1 = $4!$ ways to arrange them, or 24

Think like this:

  • For the history books (assuming we only look at the history books): 2 options for the first slot, and 1 for the last
  • For the math books (again, only look at the math books): 5 options for the first slot, $5-1=4$ for the second slot, $5-2=3$ for the third and so on
  • The same for the novels

We also have three types of books, so, the order of first-to-appear is, by the same logic, 3!

Therefore, in the the end you have $2!*5!*4!*3!=34560$ ways to arrange those books

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