Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>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
To multiply whole numbers and fractions, multiply the numerator by the whole number. Example: 1/3×4= 4/3=1 1/3
Answer:
Step-by-step explanation:
<h2>the letter in the problem is the missing number</h2>
Answer:
2
Step-by-step explanation:
An arithmetic progression with first term a and common difference d has the following first 6 terms:
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d
We are given a = 1 and a + 5d = 11.
1 + 5d = 11
5d = 10
d = 2
Answer: The common difference is 2.
40c=520
This is the proper equation because there are 40 quarters in $10. Thus, if we could only fit $10 worth of quarters in each wrapper, we can only fit 40quarters in each wrapper. C represents the amount of wrappers we will need.
Hope this helps!
