Could someone check these? basic honors algebra 1 question
of eight students in a class three will be chosen for student government positions. what is the number of permutations of those eight students to hold the three different positions?
24**
336
112
56
Can you explain how you got 24?
This permutation problem can be solved using the standard equation
P(n,r)=n!/(n-r)!
where
n=8, (8 candidates)
r=3 (three different positions out of n candidates)
Use your calculator to help you find the right answer.
Just in case, n!=1*2*3*4....*n
For example, 5!=1*2*3*4*5=120
Think about it, and you will see that your answer is way off. At the very least, the fact that the question is asking about permutations should tip you off.
There are 8 ways to pick the first position.
For each of those 8 choices, there are 7 ways to pick the next one.
Similarly, there are 6 people left for the third pick.
This gives you 8P3 ways to choose. Now what do you say?
Thank you guys. I understand now
I multiplied 8*3, I see the answer is 336!
You better either read the chapter or Google permutations !
http://betterexplained.com/articles/easy-permutations-and-combinations/
To find the number of permutations of the eight students for the three different positions, you can use the formula for permutations: nPr = n! / (n - r)!
Here, n represents the total number of students (eight), and r represents the number of positions to be filled (three).
Using this formula, you can calculate the permutations as follows:
8P3 = 8! / (8 - 3)!
= 8! / 5!
= (8 * 7 * 6 * 5!) / 5!
= (8 * 7 * 6)
= 336
Therefore, the number of permutations of the eight students to hold the three different positions is 336.