Answer: [tex]4.37\times10^3[/tex]
Step-by-step explanation:
Given : Number of choices for novels = 4
Number of choices for plays = 6
Number of choices for poetry books = 5
Number of choices for nonfiction books = 5
Total books =4+6+5+5=20
If he wants to include all 4 novels, then the number of books left to select = 9-4=5
Remaining choices for books = 20-4=16
Number of combinations of n things taking r at a time : [tex]\dfrac{n!}{r!(n-r)!}[/tex]
Then, the number of different reading schedules are possible :_
[tex]^4C_4\times^{16}C_5\\\\=\dfrac{4!}{4!(4-4)!}\times\dfrac{16!}{5!(16-5)!}\\\\=(1)\times\dfrac{16\times15\times14\times13\times12\times11!}{120\times11!}\\\\=4368=4.368\times10^3\approx4.37\times10^3[/tex]
Hence, the required answer is [tex]4.37\times10^3[/tex].
The number of different reading schedules is possible is 4368.
Given
Number of choices for novels = 4
Number of choices for plays = 6
Number of choices for poetry books = 5
Number of choices for nonfiction books = 5
Total number of books = 4+6+5+5 = 20
A combination is a way of selecting items from a collection where the order of selection does not matter.
He wants to include all 4 novels, then the number of books left to select is;
= 9 - 4 = 5
Therefore,
The number of different reading schedules is possible is;
[tex]\rm = \ ^4C_4\times ^{16}C_5\\\\= 1 \times \dfrac{16!}{(16-5)!5!}\\\\= 1\times \dfrac{16!}{11!5!}\\\\= 1\times 4368\\\\= 4368[/tex]
Hence, the number of different reading schedules is possible is 4368.
To know more about Combination click the link given below.
https://brainly.com/question/25351212
