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
Answer:
Yes
Step-by-step explanation:
What would you do if he said ok so he said yes would go?
I told him god bless him and to keep working in his vocabulary.
Answer:
B.
Step-by-step explanation:
use distributive property
4*x + 4*3 = 4x+12
If the question is what I think it is, then it's simple. All you need to do is divide 420 by 15; and you'll get the answer: 28.
Answer:
x=-3
y=3
(-3-3i)(3+5i)
Step-by-step explanation:
and we have that must equal
6-24i
3x+5y=6
5x-3y=-24 and if we work it we found out that the solutions are
x=-3 and
y=3
