Answer:
the number of possible reading schedules is 1.064301638 × 10¹²
Step-by-step explanation:
Given that :
The English teacher needs to pick 10 books to put on her reading list for the next school year.
If the English teacher picks at most 3 poetry books i.e no more than 3 poetry books from 8 books. and other books are picked from (6+4+4 ) = 14 books
Thus; the number of ways to pick the books are :
[tex]\left[\begin{array}{c}8\\0\\ \end{array}\right] \ \left[\begin{array}{c}14\\10\\ \end{array}\right]+ \left[\begin{array}{c}8\\1\\ \end{array}\right] \left[\begin{array}{c}14\\9\\ \end{array}\right] + \left[\begin{array}{c}8\\2\\ \end{array}\right] \left[\begin{array}{c}14\\8\\ \end{array}\right] + \left[\begin{array}{c}8\\3\\ \end{array}\right] \left[\begin{array}{c}14\\7 \\ \end{array}\right][/tex]
[tex]= [ \dfrac{8!}{0!(8-0)!}* \dfrac{14!}{10!(14-10!)} ] + [ \dfrac{8!}{1!(8-1)!}* \dfrac{14!}{9!(14-9)!}]+ [ \dfrac{8!}{2!(8-2)!}* \dfrac{14!}{8!(14-8)!}] + [ \dfrac{8!}{3!(8-3)!}* \dfrac{14!}{7!(14-7)!}][/tex]
[tex]= [ 1*1001]+[8*2002]+[28*3003]+[56*3432][/tex]
[tex]\mathbf{= 293293}[/tex]
However, to determine how many reading schedules that are possible we use the relation:
Number of ways to pick a book × [tex]^{10}P_{10}[/tex]
[tex]= 293293* \dfrac{10!}{(10-10)!}[/tex]
= 293293 × 10!
= 1.064301638 × 10¹²
Thus , the number of possible reading schedules is 1.064301638 × 10¹²
