SOLUTION: This question has two parts. Given a list of 6 students (i.e. Mary, Sue, Anna, Bob, Sam, John) how many different ways can they be arranged in order first to last with no repeats?

Question 651790: This question has two parts. Given a list of 6 students (i.e. Mary, Sue, Anna, Bob, Sam, John) how many different ways can they be arranged in order first to last with no repeats? I figured out that there would be 720 possible arrangements 6*5*4*3*2*1=720 different combinations. But I don't know how to figure out the second part of the question which is, "What are the number of times Anna will be listed first out of the six names?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First part is correct.

Second part:

Anna is locked in the first slot. So you have 6-1 = 5 students and 6-1 = 5 slots left.

So there are 5! = 120 ways to do this.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
This question has two parts. Given a list of 6 students (i.e. Mary, Sue, Anna, Bob, Sam, John) how many different ways can they be arranged in order first to last with no repeats? I figured out that there would be 720 possible arrangements 6*5*4*3*2*1=720.
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But I don't know how to figure out the second part of the question which is, "What are the number of times Anna will be listed first out of the six names?
Ans: 1*5*4*3*2*1 = 5! = 120 times
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Cheers,
Stan H.
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