Solve each problem part 2a. NO LINKS​

Answers:

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.

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).

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!)

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.

Problem 18

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

Add up the subtotals to get

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

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