Let's solve the problem step by step.
### Part a: Is this a permutation or combination? Explain.
This scenario is a combination. The reason is that in this problem, the order in which the students are chosen does not matter. We are only interested in which group of students is chosen, not the sequence in which they are selected. Therefore, we use combinations here.
### Part b: How many different ways can 4 students be chosen from a group of 20?
To determine the number of ways to choose 4 students out of 20, we use the formula for combinations:
[tex]\[ C(n, r) = \frac{n!}{r! \times (n-r)!} \][/tex]
where [tex]\( n \)[/tex] is the total number of items (students, in this case), and [tex]\( r \)[/tex] is the number of items to choose.
In this problem:
- [tex]\( n = 20 \)[/tex] (the total number of students)
- [tex]\( r = 4 \)[/tex] (the number of students to be chosen)
Plugging these values into the combination formula:
[tex]\[ C(20, 4) = \frac{20!}{4! \times 16!} \][/tex]
The exclamation mark "!" denotes a factorial, which is the product of all positive integers up to that number. Factorial values grow very rapidly, so let's simplify the calculations:
[tex]\[ C(20, 4) = \frac{20 \times 19 \times 18 \times 17 \times 16!}{4! \times 16!} \][/tex]
The [tex]\( 16! \)[/tex] terms cancel out from the numerator and denominator:
[tex]\[ C(20, 4) = \frac{20 \times 19 \times 18 \times 17}{4!} \][/tex]
Now calculate [tex]\( 4! \)[/tex] (which is [tex]\( 4 \times 3 \times 2 \times 1 \)[/tex]):
[tex]\[ 4! = 24 \][/tex]
So the calculation becomes:
[tex]\[ C(20, 4) = \frac{20 \times 19 \times 18 \times 17}{24} \][/tex]
Performing the multiplication in the numerator:
[tex]\[ 20 \times 19 = 380 \][/tex]
[tex]\[ 380 \times 18 = 6840 \][/tex]
[tex]\[ 6840 \times 17 = 116280 \][/tex]
Now divide by the denominator:
[tex]\[ C(20, 4) = \frac{116280}{24} = 4845 \][/tex]
Thus, the number of different ways to choose 4 students from a group of 20 is:
[tex]\[ \boxed{4845} \][/tex]
