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:
a numerical or constant quantity placed before and multiplying the variable in an algebraic expression
Step-by-step explanation:
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Answer:
not sure i it is one of these answers but if so , it d
Step-by-step explanation:
Answer with explanation:
The given statement is which we have to prove by the principal of Mathematical Induction
1.→For, n=1
L H S =2
R H S=1
2>1
L H S> R H S
So,the Statement is true for , n=1.
2.⇒Let the statement is true for, n=k.
---------------------------------------(1)
3⇒Now, we will prove that the mathematical statement is true for, n=k+1.
Hence it is true for, n=k+1.
So,we have proved the statement with the help of mathematical Induction, which is
Answer: 18/7
Step-by-step explanation: random.
