Please help with this one question from my homework please i need it quickly

Question

Question 1: Eight different coffee mugs have to be arranged in a showcase. How many different arrangements are possible?

A.) 50,000

B.) 40,320

C.) 30,760

D.) 40,000

 

Question 2: Tina has 20 rings. In how many ways can she select a set of 3 rings?

A.) 1000

B.) 850

C.) 950

D.) 1140

Answer 1

Question 1:

For this problem, we will use a permutation because we are arranging the mugs, and the order matters. A permutation is used when changing the order creates a different result.

For the first position, we can choose any of the 8 mugs so we have 8 choices. After choosing the first mug, only 7 mugs remain for the second position, giving us 8 x 7 choices. After choosing the second mug, 6 mugs remain for the third position, and this pattern continues. So, we can do the factorial of 8 to find the total number of possible arrangements:

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

= 40,320 arrangements are possible

Question 2:

This question is different from the first one because Tina is selecting a group of rings. She is not arranging them in a particular order. Therefore, we use a combination. A combination is used when the order does not matter.

The combination formula is:

ⁿC_r = C(n, r) = n! / r!(n-r)!

where:

1. n - the total number of items in the set
2. r - the number of items you want to choose
3. ! - factorial

So, let's plug in our numbers for this problem into the combination formula:

²⁰C₃ = 20! / (3! x 17!) We can cancel 17! from numerator and denominator

= (20 x 19 x 18) / (3 x 2 x 1)

= 6840 / 6

= 1,140 ways Tina can select three rings

Answer 2

Quation 1. P(8) = 8! = 40320. Answer B.)

Quation2. C(20, 3) = 20·19·18/(1·2·3) = 1140 Answer D).