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February 23rd, 2014, 02:04 PM   #1
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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.
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February 23rd, 2014, 06:09 PM   #2
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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.

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February 24th, 2014, 08:15 AM   #3
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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?
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February 24th, 2014, 08:17 AM   #4
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Re: Probability conundrum

Is it.... 11! 2!2!
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February 24th, 2014, 10:01 AM   #5
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Re: Probability conundrum

Yes.
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