A board of directors consisting of eight members is to be chosen from a pool of 28 candidates. The board is to have a chairman, a treasurer, a secretary, and five other members. In how many ways can the board of directors be chosen?
For permutations without repetitions, the formula is n!/((n-r)!), where n is the number of things to choose from, and you choose r of them There is no repetition, and order matters.
Think about it this way. You have 28 candidates total. For the position of chairman, there are 28 possible people. After you select one person for chairmen, there are 27 people left - 27 possible people.
With each position that is filled, you have one less person for the job. At the end, you'll have all 8 positions filled, and 20 people left over.
So then we have 28!/((28-8)!) = 28x27x26x25x24x23x22x21. That should give you your answer.
To find the number of ways the board of directors can be chosen, we need to consider the positions of the chairman, treasurer, and secretary individually, as well as the other five members.
1. Choosing the chairman: We have 28 candidates, and we need to select 1 for the position of chairman. So, there are 28 choices for the chairman.
2. Choosing the treasurer: After choosing the chairman, we are left with 27 candidates for the treasurer position. So, there are 27 choices for the treasurer.
3. Choosing the secretary: After choosing the chairman and treasurer, we are left with 26 candidates for the secretary position. So, there are 26 choices for the secretary.
4. Choosing the other five members: After choosing the chairman, treasurer, and secretary, we are left with 25 candidates for the remaining five positions. Since the positions do not have any specific order, we can choose the five members from the remaining candidates in 25C5 ways (combination formula).
The expression 25C5 represents the number of ways to choose 5 members from a group of 25 candidates, which is calculated using the combination formula:
nCr = n! / (r!(n - r)!)
In this case, n = 25 and r = 5. Plugging these values into the formula, we get:
25C5 = 25! / (5!(25-5)!)
= 25! / (5!20!)
Simplifying further:
25! = 25 * 24 * 23 * ... * 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
20! = 20 * 19 * 18 * ... * 3 * 2 * 1
Canceling out common terms:
25! / (5!20!) = (25 * 24 * 23 * ... * 3 * 2 * 1) / (5 * 4 * 3 * 2 * 1 * 20 * 19 * 18 * ... * 3 * 2 * 1)
= (25 * 24 * 23 * ... * 21) / (5 * 4 * 3 * 2 * 1)
= 53,130
Therefore, there are 53,130 ways to choose the other five members.
To find the total number of ways to choose the entire board of directors, we multiply the number of choices from each step:
Total number of ways = Number of choices for chairman * Number of choices for treasurer * Number of choices for secretary * Number of choices for the other five members
Total number of ways = 28 * 27 * 26 * 53,130
= 798,336 * 53,130
= 42,414,566,080
Therefore, there are 42,414,566,080 ways to choose the board of directors.