Decide if it's permuatation or combination .
1-how many teams of 4 horses be made if there were 9 horses in total ?
2-mike has nice baseball trophies to arrange on the shelf . How many different ways can be arranged ?
3-in a math class , there are 24 students,the teacher picks 4 students to help do a demonstration.how many diffrent groups of 4 could she have chosen ?
if order matters, it's permutations
if not, then it's combinations
To determine if it's a permutation or combination problem, we need to consider whether the order of selection matters or not.
1. How many teams of 4 horses can be made if there were 9 horses in total?
Since the order doesn't matter when selecting the horses for a team, this is a combination problem. We can use the formula for combinations, which is nCr, where n is the total number of horses and r is the number of horses being chosen. In this case, we want to calculate 9C4.
The combination formula is:
nCr = n! / [(n-r)! * r!]
Calculating 9C4:
9! / [(9-4)! * 4!] = 9! / (5! * 4!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126
Therefore, there are 126 different teams of 4 horses that can be made.
2. How many different ways can Mike arrange his baseball trophies on the shelf?
Since the order of arrangement matters, this is a permutation problem. We can use the formula for permutations, which is nPr, where n is the total number of trophies and r is the number of trophies being arranged. In this case, we want to calculate P(n, n), where n is the number of trophies Mike has.
The permutation formula is:
nPr = n!
Calculating P(n, n):
P(n, n) = n! = n!
Therefore, there are n! (n factorial) different ways Mike can arrange his trophies on the shelf.
3. How many different groups of 4 students could the teacher have chosen from a class of 24 students?
Since the order doesn't matter when selecting the students for the group, this is a combination problem. We can again use the combination formula, which is nCr, where n is the total number of students and r is the number of students being chosen. In this case, we want to calculate 24C4.
Calculating 24C4:
24! / [(24-4)! * 4!] = 24! / (20! * 4!) = (24 * 23 * 22 * 21) / (4 * 3 * 2 * 1) = 10,626
Therefore, there are 10,626 different groups of 4 students that the teacher could have chosen.