An English teacher needs to pick 10 books to put on his reading list for the next school year, and he needs to plan the order in which they should be read. He has narrowed down his choices to 5 novels, 6 plays, 5 poetry books, and 6 nonfiction books.

Step 1 of 2 : If he wants to include no more than 4 plays, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.

I know it starts with 6c0 x ___ + 6c1 x ____ = 6c2 x ______ + 6c3 x _______ + 6c4

An English teacher needs to pick 10 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 5 novels, 5 plays, 3 poetry books, and 5 nonfiction books.

Well, that's a lot of "c"'s and underscores! Let's break it down and make it a bit more fun.

Step 1: Choose 0 plays out of the 6 available plays. Easy peasy! So we have 6C0 (read as "6 choose 0") which is 1.

Step 2: Choose 1 play out of the remaining 6. We have 6C1, which is also 6.

Step 3: Choose 2 plays out of the remaining 6. We have 6C2. To calculate this, we can use the formula: 6! / (2! * (6-2)!), where "!" represents factorial. If we plug in the numbers, we get 6! / (2! * 4!), which simplifies to (6 * 5) / (2 * 1), resulting in 15.

Step 4: Choose 3 plays out of the remaining 6. We have 6C3, which is 20.

Step 5: Choose 4 plays out of the remaining 6. We have 6C4, which is 15.

Now, we add up all the possibilities: 6C0 + 6C1 + 6C2 + 6C3 + 6C4 = 1 + 6 + 15 + 20 + 15 = 57.

So, there are 57 different reading schedules possible. In scientific notation rounded to the hundredths place, that would be 5.70 x 10^1.

To solve this problem, we need to find the number of different reading schedules possible by including no more than 4 plays.

The teacher has narrowed down the choices to 5 novels, 6 plays, 5 poetry books, and 6 nonfiction books.

Step 1: Calculating the number of schedules with 0, 1, 2, 3, and 4 plays:

- For 0 plays: We need to choose 10 books (since the teacher wants to pick 10 books), and we have 5 novels, 5 poetry books, and 6 nonfiction books to choose from. So the number of schedules with 0 plays is obtained by multiplying the number of choices for each book type: 5 novels x 5 poetry books x 6 nonfiction books = 150 schedules.

- For 1 play: We need to choose 9 more books (since we already chose 1 play) from the remaining options. We have 5 novels, 5 poetry books, and 6 nonfiction books, and we need to choose 2 more book types (since we already have 1 play). We can choose these 2 book types in (6 choose 2) = 15 ways. So the number of schedules with 1 play is obtained by multiplying the number of choices for each component: 5 novels x 5 poetry books x 15 choices for the 2 additional book types x 2 choices for the order of the chosen book types (novels, poetry) = 1500 schedules.

- Similarly, for 2 plays: We need to choose 8 more books (since we already chose 2 plays) from the remaining options. We can choose these 3 book types in (6 choose 3) = 20 ways. So the number of schedules with 2 plays is obtained by multiplying the number of choices for each component: 5 novels x 5 poetry books x 20 choices for the 3 additional book types x 3 choices for the order of the chosen book types (novels, poetry, plays) = 15000 schedules.

- For 3 plays: We need to choose 7 more books (since we already chose 3 plays) from the remaining options. We can choose these 4 book types in (6 choose 4) = 15 ways. So the number of schedules with 3 plays is obtained by multiplying the number of choices for each component: 5 novels x 5 poetry books x 15 choices for the 4 additional book types x 4 choices for the order of the chosen book types (novels, poetry, plays, nonfiction) = 15000 schedules.

- For 4 plays: We need to choose 6 more books (since we already chose 4 plays) from the remaining options. We can choose these 4 book types in (6 choose 4) = 15 ways. So the number of schedules with 4 plays is obtained by multiplying the number of choices for each component: 5 novels x 5 poetry books x 15 choices for the 4 additional book types x 5 choices for the order of the chosen book types (novels, poetry, plays, nonfiction) = 37500 schedules.

Step 2: Summing up the schedules:

To find the total number of schedules, we need to add up the number of schedules obtained in each step:

Total number of schedules = number of schedules with 0 plays + number of schedules with 1 play + number of schedules with 2 plays + number of schedules with 3 plays + number of schedules with 4 plays.

Total number of schedules = 150 + 1500 + 15000 + 15000 + 37500.

Total number of schedules = 69000.

Since we need to express the answer in scientific notation rounding to the hundredths place, the answer is approximately 6.90 x 10^4.

Therefore, there are approximately 6.90 x 10^4 different reading schedules possible if the teacher wants to include no more than 4 plays.

number of ways to choose 10 from 22

= C(22,10)
what can't we have:
choosing 5 plays + choosing 6 plays
= C(6,5)xC(16,5) + C(6,6)xC(16,4)

number of ways = C(22,10) - ( C(6,5)xC(16,5) + C(6,6)xC(16,4) )
= 646646 - (26208+1820)
= 618618
or in scientific: 6.19x10^5 (but why write it that way)

long way:
C(6,0)xC(16,10) + C(6,1)xC(16,9) + ... + C(6,4)xC(16,6)
= 618618