An english teacher needs to pick 9 books to put on her reading list for the next school year, and she needs to plan the order in which they should be read. she has narrowed down her choices to 7 novels, 3 plays, 4 poetry books, and 4 nonfiction books. step 1 of 2 : if she wants to include no more than 3 novels, how many different reading schedules are possible? express your answer in scientific notation rounding to the hundredths place.

Hi, so you have 7 novels, 3 plays, 4 poetry books and 4 nonfiction, and rounding to the hundredths is rounding to the second number after the decimal point.

the teacher must chose 9 of them, and the restriction is no more than 3 novels. Because the restriction is only in the novels, all the other types are the same in our eyes, this is:

we have 7 novels, and 11 others.

if we put 3 novels.

we have 3 sockets where in the first socket we have 7 options, in the second 6, and in the third 5. this is 7*6*5.

and we have 6 sockets (for the other books) where in the first we have 11 options, in the second 10, and so on. So the total combinations are 11*10*9*8*7*6.

combined are: (7*6*5)(11*10*9*8*7*6) = 210*332640=69854400= (there are a lot of combinations)

if we put two novels, the calculus are similar to before; we have 7*6 combinations for the novels, and 11*10*9*8*7*6*5 combinations for the others, combinating them we get: (7*6)*(11*10*9*8*7*6*5) which is the same number as before, this time the "5" is in the right parenthesis.

then for two novels we got combinations.

And for 1 novel:

For the novels we have 7 combinations, and for the others we have:

11*10*9*8*7*6*5*4, and combinating them we get:

7*(11*9*8*7*6*4*5*4) = 46569600 =

for 0 novels we get only combinations of the other 11 books, so the total combinations are : 11!/2! = 11*10*9*8*7*6*5*4*3 = 19958400 =

So the total combinations will be the sum of the combinations for 3 novels, for 2 novels, for 1 and for 0 novels, this is:

N =  

N = = is the total number of reading schedules.

i would be happy to . most people on would also love to i'm sure. if you just want someone who has done algebra before and want a little more personal relationship than just begging strangers for , i can certainly be of service. i do have college myself so i might not be 24/7.