In how many ways can 4 students from a group of 9 be seated in a row of 4 chairs?
I tried 9C4 = 126, 126C4
am I going the right direction?
thanks in advance
You have chair nr. 1, chair nr.2, chair nr. 3 and chair number 4. If you seat a student at chair nr. 1 first, then at chair nr. 2, etc., then you have 9 choices for to choose the student who will sit in chair nr. 1, then 8 choices for who to seat in chair nr. 2 etc. etc., so the total number of posibilities is:
9*8*7*6
You can also say that there are
Binomial[9,4] ways to choose 4 students out of 9. And then there are 4! ways to choose a seating order for any given choice of the students:
Binomial[9,4] 4! = 9!/5! = 9*8*7*6
I understand now, thanks!
Yes, you are on the right track! To solve this problem, you need to use the concept of combinations.
The initial step you took, 9C4, is correct. It represents the number of ways to choose 4 students out of a total of 9 students. This gives you the number of different groups that can be formed with 4 students.
However, you are looking for the number of ways these 4 students can be seated in a row of 4 chairs. To calculate this, you need to consider the concept of permutations.
To find the number of ways to arrange the 4 selected students in a row of 4 chairs, you can use the formula for permutations. The formula is nPr = n! / (n-r)!, where n represents the total number of objects and r represents the number of objects to be selected.
In this case, you have selected 4 students out of 9, so you can calculate the number of ways to arrange them using 4P4.
Using the formula, 4P4 = 4! / (4-4)! = 4! / 0! = 4! = 4 x 3 x 2 x 1 = 24.
Therefore, there are 24 different ways to seat 4 students from a group of 9 in a row of 4 chairs.
Please note that the calculation for 126C4 that you mentioned is not applicable in this situation because it represents the total number of possible combinations of 4 students out of a group of 126, which is unrelated to the given problem.