In a Mathematics class with 30 students, the teacher wants 2 different ways to present the solutions to problems 3 and 5 on the board. In how many ways can be teacher assign the
problems?
To find the number of ways the teacher can assign the problems, we need to consider that there are 30 students and 2 problems to assign.
First, let's find the number of ways to select the first problem to be presented. Since there are 2 problems, there are two options for the teacher to choose the first problem.
Now, for the second problem, since the teacher wanted two different ways to present it, there are 29 students left after assigning the first problem. So there are 29 options for the second problem.
Therefore, the total number of ways the teacher can assign the problems is calculated by multiplying the number of options for each problem: 2 options for the first problem multiplied by 29 options for the second problem.
2 * 29 = 58
Hence, the teacher can assign the problems in 58 different ways.
To solve this problem, we can use the concept of permutations.
First, we need to determine the number of ways the teacher can select two problems out of the five given (problems 3 and 5). This can be found using the combination formula:
nCk = n! / (k!(n-k)!)
n = 5 (total number of problems)
k = 2 (number of problems to be selected)
So, we have:
5C2 = 5! / (2!(5-2)!)
= 5! / (2!3!)
= (5 * 4 * 3!) / (2!3!)
= (5 * 4) / 2
= 10
Therefore, there are 10 different ways the teacher can select two problems (3 and 5) out of the given five.
Now that the teacher has selected two problems, each problem can be assigned to either presentation 1 or presentation 2. So, for each selected pair of problems, there are 2 ways to assign them to the presentations.
Since there are 10 ways to select two problems, we can multiply this number by 2 to get the total number of ways to assign the problems:
10 * 2 = 20
Therefore, the teacher can assign the problems in 20 different ways.