How many ways can a teacher arrange four students in the front row with a total of 30 students?
I'm confused
Is this a combination or a permutation? What do you think?
Idk
Permutation?
Does the order in which the four students placed in the front row matter?
Yes?
So let's say that Bill, Jim, Sue, and Betty are chosen to sit in the front row. Does it matter in what order they sit?
No
Great. So, as long as those four are chosen, it doesn't matter in which order they're chosen. If the order doesn't matter, is that a combination or a permutation?
Permutation?
Sorry, no. If the order doesn't matter, then it's a combination. If the order does matter, then it's a permutation. So this question is about calculating a combination. Do you know how to do that?
No
The number of combinations of choosing r items out of a set on n items is given by\[_{n}C _{r}=\frac{ n! }{ r!\left( n-r \right)! }\]Do you understand this equation?
No
OK. In this question, n is the total number of people. How many is that?
30
Yup. And r is the number of people that are chosen to sit in the front row. How many is that?
4
Right. So the calculation becomes\[_{30}C _{4}=\frac{ 30! }{ 4!\left( 30-4 \right)! }=\frac{ 30! }{ 4!26! }\] Can you calculate this answer?
Not really
Do you know what the factorial symbol (!) means?
4!x26! ?
Yes
I'm using a scientific calculator
Great. So you have\[_{30}C _{4}=\frac{ 30\times29\times28\times27\times26\times25\times24\times...\times2\times1 }{\left( 4\times3\times2\times1 \right)\left( 26\times25\times24\times...\times2\times1 \right) }\]Make sense?
If so, there are a lot of common factors in the numerator and denominator that cancel out.
Still there?
Yes sorry had to do something
Cancel out the common factors and calculate the result. What do you get?
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