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Math Probability question...

Originally posted by: Excelsior
Obviously we are assuming that they have different names.

Already tried 1/120

You already tried 1/120?

Already tried what?

The chance that A will sit in the 1st is 1/5. The chance that B will sit in the next seat is 1/4. The chance that C will sit in the next seat is 1/3. The chance that D will sit in the next seat is 1/2.

5*4*3*2=120

Edit: Can they all sit down at the same time?
 
Originally posted by: Chaotic42
Originally posted by: Excelsior
Obviously we are assuming that they have different names.

Already tried 1/120

You already tried 1/120?

Already tried what?

The chance that A will sit in the 1st is 1/5. The chance that B will sit in the next seat is 1/4. The chance that C will sit in the next seat is 1/3. The chance that D will sit in the next seat is 1/2.

5*4*3*2=120

Edit: Can they all sit down at the same time?

No, it IS 120, you are right. The girl wasn't putting it in right. I thought it was 120 too (she was doing decimal form instead of fractions).
 
Originally posted by: Excelsior
No, it IS 120, you are right. The girl wasn't putting it in right. I thought it was 120 too (she was doing decimal form instead of fractions).
Ok, cool. I was wondering what I was doing wrong!

:beer:
 
assuming their names are all different its just 5 factorial ways.


i suppose if the chairs are in a row, there are two ways they can sit in order


like

abcde

or edcba


so thats 2/5! 1/60
 
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