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Why can I have more than two 2!'s when Im dividing a #? Ex. 26800/(2!)(2!)(2!) <- Wont let me add the third 2!

Guest Sep 26, 2017

Best Answer 

 #2
avatar+26814 
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It only likes one implied multiplication.  You have to enter it as 26800/(2!)(2!*2!)

Alan  Sep 26, 2017
 #1
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If you are using the key labeled "nPr", it only takes  TWO parameters. Example: nPr(52, 5), which means this: 52P5= 52 x 51 x 50 x 49 x 48 = 52! / (52 -5)! = 311,875,200

Guest Sep 26, 2017
 #2
avatar+26814 
+3
Best Answer

It only likes one implied multiplication.  You have to enter it as 26800/(2!)(2!*2!)

Alan  Sep 26, 2017

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