Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students do not belong to the same club. We will see that there are 110 different ways in which 2 students from different clubs can be selected.

So there are 3 clubs:

Club A, with 10 students.
Club B, with 4 students.
Club C, with 5 students.

The possible combinations of 2 students from different clubs are

Club A with club B
Club A with club C
Club B with club C.

The number of combinations for each of these is given by the product between the number of students in the club, so we get:

Club A with club B: 10*4 = 40
Club A with club C: 10*5 = 50
Club B with club C. 4*5 = 20

For a total of 40 + 50 + 20 = 110 different combinations.

This means that there are 110 different ways in which 2 students from different clubs can be selected.

If you want to learn more about combination and selections, you can read:

brainly.com/question/251701

6 0

2 years ago

You are making bouquets to sell. You have 60 roses, 84 daisies, and 24 lilies. Each bouquet will have the same number of each ki

laiz [17]

You can make 6 bouquets. Each bouquet will have 10 roses, 14 daisies, and 4 lilies.
The math:
FIND THE GREATEST COMMON FACTOR
60/6=10--->10 ROSES
84/6=14--->14 DAISIES
24/6=4--->4 LILIES

Hope this helped!

7 0

3 years ago

Ms. Kincaid keeps a supply of dimes and quarters in her car to pay for highway tolls. A week’s supply of toll coins contains 5 m

Alecsey [184]

Answer: (A) 10

Step-by-step explanation:

Value Quantity = TOTAL Value

dimes: .10 Q + 5 = .10(Q = 5)

quarters: .25 Q = .25Q

Dimes + Quarters = $4.00

.10(Q + 5) + .25Q = 4.00

.10Q + .50 + .25Q = 4.00

.50 + .35Q = 4.00

.35Q = 3.50

Q = 10

Quarters = 10

Dimes = Q + 5

= 10 + 5

= 15

8 0

3 years ago

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