Question

Brett M.-Week 5 Brett Mize posted Jan 31, 2025 9:57 AM Subscribe Good morning everyone, 1. There is a club of 15 people. There are five positions in the governmental structure of the club, The President, Vice President, the Secretary, the treasurer, and the head of communications for the club. No person can have more than one position within the club government structure. 2. How many different potential makeups of club government can be made up with each person only being able to hold one position in the government makeup? 3. This problem can be solved with the equation for solving r-permutations with \( n \) elements, \( p(n, r)=n!/(n-r)!.15 \times 14 \times 13 \times 12 \times 11 \times 10!/ 10! \) the factorials of 10 cancel each other out through the division. This leaves the equation \( 15 \times 14 \times 13 \times 12 \times 11=3,030,720 \) combinations. 4. In a club built up of 15 people, with five government positions available. In a situation where each club member can only hold one position in government, there are \( 3,030,720 \) unique potential governmental makeups. 5. As a member of the club this can help to know how many different ways the government can be built, or at least let you know that there are many options. It is also interesting just from a mathematical standpoint. -Brett

          Brett M.-Week 5
Brett Mize posted Jan 31, 2025 9:57 AM
Subscribe

Good morning everyone,
1. There is a club of 15 people. There are five positions in the governmental structure of the club, The President, Vice President, the Secretary, the treasurer, and the head of communications for the club. No person can have more than one position within the club government structure.
2. How many different potential makeups of club government can be made up with each person only being able to hold one position in the government makeup?
3. This problem can be solved with the equation for solving r-permutations with \( n \) elements, \( p(n, r)=n!/(n-r)!.15 \times 14 \times 13 \times 12 \times 11 \times 10!/ 10! \) the factorials of 10 cancel each other out through the division. This leaves the equation \( 15 \times 14 \times 13 \times 12 \times 11=3,030,720 \) combinations.
4. In a club built up of 15 people, with five government positions available. In a situation where each club member can only hold one position in government, there are \( 3,030,720 \) unique potential governmental makeups.
5. As a member of the club this can help to know how many different ways the government can be built, or at least let you know that there are many options. It is also interesting just from a mathematical standpoint.
-Brett

Added by Gregg M.

Question

Please give Ace some feedback