ways to ans. 5 Qs out of 12, given you must ans. last Q

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stacey

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May 6, 2007
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I used Combinations for this question:

Select five questions from an exam containing 12 questions if you must answer the last question?

12 C 5 = 792
 
That really hurts.

If you would answer my question, you may change your opinion.
 
I thought I had answered your question I apologize for posting the same question under two post. My mistake. I tried 12 I tried 12 with comb and 12 with premutation. I don't think that is how you get the answer. I am wonder if it is a binomial? The hardest thing I have with statistic is the wording. If the question was straight forward it would be easier. I did the 11 instead of the 12 as to the last wording in the question? My instrucotr still has not answered the question either on the discussion board.
Stacey
 
Stacey, do to the "5" what you did to the "12" :idea:
 
I tried that it comes to zero. It wouldn't work with 5C12 or 5P12
 
> I did the 11 instead of the 12 as to the last wording in the question?

You took 1 away from the "12", right?
I said do the same to the "5"; take 1 away: so 11C4 : kapish?

RELAX and think about it:
12 and 5, but last must be answered;
so you answer the last; what's left? 4 out of 11, right ? :idea:

ALSO, remember that with xCy, x must be = or greater than y;
you can't Choose 12 from 5, right?
xCy reads "from x, Choose y".
 
stacey said:
I tried that it comes to zero. It wouldn't work with 5C12 or 5P12
Click to expand...
I'm really curious to know what it was you did that came to zero (0). Can you demonstrate?
 
Denis,
Thanks. I really like this but again it is the wording. Moving right along doing Chapt 6 Normal Distributions. No homework is given just read, do the video lectures and take quizes or exams. I am going to work thru the questions at the end of chapt 6. I am using the calculator but stumbling along.
Happy Canadian day to you!!!!
Stacey
 
Welcome, Stacey...keep plugging...

TK, I think Stacey tried 5C11 and 5C12, so 0 was returned.
 
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