Question

The math club is electing new officers. There are 5 candidates for president, 6 candidates for vice-president, 2 candidates for secretary, and 3 candidates for treasurer. How many different combinations of officers are possible?

Answer 1

180

Explanation:

We are going to solve for the individual offices.

For president,there are 5 candidates. The number of ways of choosing one candidate is

5!(1!/4!).What 5! means is 5×4×3×2×1 and the sample applies for 1! and 4!

Solving for the presidential candidate, we have 5!(1!/4!)= 5×4×3×2×1=120 divided by

4×3×2×1=24 which then equals 5.

Follow the same procedure for the other offices.

For vice president, there are 6 candidates.The number of ways of choosing one candidate is

6!(1!/5!)

For secretary,there are 2 candidates.The number of ways of choosing one candidate is

2!(1!/1!)

And for treasurer, the number of ways of choosing a candidate is

3!(1!/2!)

At the end of the day,we have 5,6,2 and 3 respectively.

Multiply them all and we have 180 ways

Answer 2

180 different combinations

Explanation:

To find the number of different combinations possible, We first find the number of possibilities for each place, and then multiply all possibilities.

Number of possibilities for president: 5

Number of possibilities for vice-president: 6

Number of possibilities for secretary: 2

Number of possibilities for treasurer: 3

Number of different combinations: 5*6*2*3 = 180

(We can form 180 different groups of 1 president, 1 vice-president, 1 secretary and 1 treasurer)