Math Is Fun Forum

RJay

Member

Registered: 2012-02-06

Posts: 2

Permutations............................

Please help this simple kid from Africa. I have a question relating to an assessment - there are 4 questions in the assessment. In order not to repeat questions in an assessment, question 1 could be any of 13 questions, question 2 any of 21 questions, question 3 any one of 14 questions and question 4 any one of 12 questions. PLEASE, HOW DO I CALCULATE THE NUMBER OF POSSIBLE PERMUTATIONS FOR THE ASSESSMENT??
Your help would be GREATLY appreciated...........

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Permutations............................

Hi RJay;

Welcome to the forum.

All you have to do is mutiply by the number of questions. That is provided that there is no intersection between the 13, 21,14 and 12 questions.

13 * 21 * 14 * 12 = 45864

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

RJay

Member

Registered: 2012-02-06

Posts: 2

Re: Permutations............................

Thank you very much for your input!!!!!!!!!!!!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Permutations............................

Hi RJay;

You are welcome.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.