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February 23rd, 2014, 02:04 PM  #1 
Newbie Joined: Feb 2014 Posts: 3 Thanks: 0  Probability conundrum
Hi guys. Im struggling to make sense of a question I have encountered recently and am relying on some expertise to help me out: The letters of the word P R O B A B I L I T Y are written on 11 cards. These cards are then arranged on a table 1.How many arangements will start with a vowel 2.How many arrangements will end in a B 3.how many arrangements will start with a vowel and end in a B Now, I have come to my own conclusions but my problem particularly lies with the Bs. Does the fact that the 2 Bs are the exact same play a role when calculating? My answer to 1: 14,515200 2: 7,257,600 3: 2,903,040 The last one i worked out as follows: 4x9x8x7x6x5x4x3x2x1x2. The last 2 is because there are 2 Bs. But should this be a 1 because both Bs are identical, therefore it does not matter which one takes the last place. Any help at all would be much appreciated as I am struggling. Thanks a lot guys. 
February 23rd, 2014, 06:09 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Re: Probability conundrum
For each pair of identical letters, divide out 2! (where applicable): 1. (the number of arrangements starting with O or A + the number of arrangements starting with I), 2. 3. 
February 24th, 2014, 08:15 AM  #3 
Newbie Joined: Feb 2014 Posts: 3 Thanks: 0  Re: Probability conundrum
Thanks a lot! How do i go about the more simple route of finding the number of arrangements that PROBABILITY can be arranged then? 
February 24th, 2014, 08:17 AM  #4 
Newbie Joined: Feb 2014 Posts: 3 Thanks: 0  Re: Probability conundrum
Is it.... 11! ÷ 2!2!

February 24th, 2014, 10:01 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Re: Probability conundrum
Yes.


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