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Solve each problem part 2a. NO LINKS!!!!​

Solve each problem part 2a. NO LINKS!!!!​-example-1
User Hce
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Answers:

  • Problem 14) There are 1093 different ice creams possible.
  • Problem 15) There are 120 ways to pick the reading list.
  • Problem 16) We have 1365 combinations.
  • Problem 17) There are 315 different combos of cars possible.
  • Problem 18) You can go on 21,700 different combinations of rides.

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Explanations:

Problem 14

There are n = 10 toppings to choose from.

If you order r = 0 toppings, then there is nCr = 13C0 = 1 way to do this.

Use the combination formula nCr = (n!)/(r!*(n-r)!)

If you order r = 1 topping, then there are 13C1 = 13 ways to do this.

If r = 2, then we have 13C2 = 78 different combos

If r = 3, then 13C3 = 286

If r = 4, then 13C4 = 715

Add up those subtotals: 1+13+78+286+715 = 1093

Note the use of nCr instead of nPr. The order of toppings doesn't matter.

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Problem 15

Again order doesn't matter. We have n = 10 books to pick from and r = 3 selections to make.

nCr = (n!)/(r!*(n-r)!)

10C3 = (10!)/(3!*(10-3)!)

10C3 = (10*9*8*7!)/(3!*7!)

10C3 = (10*9*8)/(3!)

10C3 = (10*9*8)/(3*2*1)

10C3 = (720)/(6)

10C3 = 120

There are 120 different ways to pick the three books (from a pool of ten).

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Problem 16

Follow the same steps as problem 15. This time you'll have n = 15 and r = 4. The answer you should get is 1365

Or you could follow these steps below:

We have 15*14*13*12 = 32,760 permutations and 4! = 24 different ways to arrange any group of 4, so we have (32,760)/(24) = 1365 combinations.

nCr = (nPr)/(r!)

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Problem 17

There are 7C1 = 7 ways to pick the color and 10C2 = 45 ways to pick the two options. Therefore, there are 7*45 = 315 different cars possible.

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Problem 18

"At least 15" means "15 or more".

  • If you ride on r = 15 rides, out of n = 20 total, then we have nCr = 20C15 = 15504 different combinations.
  • If you ride on r = 16 rides, out of n = 20 total, then we have nCr = 20C16 = 4845 different combinations.
  • If you ride on r = 17 rides, out of n = 20 total, then we have nCr = 20C17 = 1140 different combinations.
  • If you ride on r = 18 rides, out of n = 20 total, then we have nCr = 20C18 = 190 different combinations.
  • If you ride on r = 19 rides, out of n = 20 total, then we have nCr = 20C19 = 20 different combinations.
  • If you ride on r = 20 rides, out of n = 20 total, then we have nCr = 20C20 = 1 combo only

Add up the subtotals to get

15504+4845+1140+190+20+1 = 21,700

User Gohmz
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