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Perms and combs question (1 Viewer)

rking24

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Hey guys came across a question which i couldn't figure out, would really appreciate if someone can help me out

how many different arrangements are possible if 3 letters are randomly selected from the word challenge and arranged into 'words'?
 

trecex1

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Hey guys came across a question which i couldn't figure out, would really appreciate if someone can help me out

how many different arrangements are possible if 3 letters are randomly selected from the word challenge and arranged into 'words'?
Split it into two cases
1) A pair is selected e.g two l's
2) No pair is selected (3 single letters)
 

BenHowe

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With these question just write out the word but in terms of the occurrences of the letters.

1c,1h,1a,2l,2e,1n,1g

PatternSelectArrangeTotal
x+y+z7c33!7c3x3!
2x+y2c1x6c13!/(2!1!)2c1x6c1x3!/(2!1!)

Summing the totals gives 246

Can you re-write the total of the first line using a different combinatoric notation? If so, why?

Note it would not be then valid to do a probability question, based on the number of outcomes listed above, since the outcomes arn't equally likely. For example: You would get 2 cases for the l's instead of 1 for the h etc.

You would need to regard the letters as distinguishable to produce equally likely outcomes, then you could evaluate the probability. To check here's a question. What's the probability that:

a) You select 3 vowels, one at a time
bi) You select 2 vowels and a consonant, simultaneously
bii) You select 2 vowels and a consonant one at a time
biii) You select 2 vowels and a consonant with repetition
 
Last edited:

rking24

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I had the exact same thought but turns out to be wrong.. damn this thing confuses me

Thanks anyways:)
 

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