A class elects two officers, a president and a secretary/
treasurer, from its 12 members. How many different
ways can these two offices be filled from the members of
the class?
Fill the president position in 12 different ways, leaving 11 to pick from for the secretary.
So, number of ways = 12x11 = 132
In order to select new board members, the French club held an election. 3 out of the 15 members of the French club voted in the election. What percentage of the members voted?
To find the number of different ways to fill the two offices (president and secretary/treasurer) from the 12 members of the class, we can use the concept of combinations.
In this case, we want to find the number of different combinations of 12 members taken 2 at a time, as order doesn't matter for these positions.
The formula to calculate combinations is:
C(n, r) = n! / (r!(n - r)!)
Where n is the total number of items, and r is the number of items we want to select.
For this problem, we have:
n = 12 (12 members in the class)
r = 2 (selecting 2 officers)
Plugging these values into the formula, we get:
C(12, 2) = 12! / (2!(12 - 2)!)
= 12! / (2!10!)
Now let's calculate the factorial values:
12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
2! = 2 × 1
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Substituting these values back into our formula:
C(12, 2) = (12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / ((2 × 1) × (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
Now, we can simplify the equation:
C(12, 2) = (12 × 11) / (2 × 1)
Calculating further:
C(12, 2) = 132 / 2
= 66
Therefore, there are 66 different ways to fill the two offices from the 12 members of the class.