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anonymous

  • 5 years ago

so, we have got these letters abcad. how many groups with 4 letters can we create, so that we do not repeat the letters?? I think there might be 4!/2 groups !

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  1. anonymous
    • 5 years ago
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    \[{5\choose 1} * {4 \choose 1}*{3 \choose 1}*{2 \choose 1} = 5!\]

  2. anonymous
    • 5 years ago
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    I do not understand? as the letter a is two times we should divide with 2! is it right?

  3. anonymous
    • 5 years ago
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    Ack! I didn't notice a was in there twice.

  4. anonymous
    • 5 years ago
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    If we do not repeat and the a's are indistinguishable there's only 1 way. Because choosing one a is the same as choosing the other we are really only choosing 4 letters from a collection of 4 letters.

  5. anonymous
    • 5 years ago
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    Or did you mean groups with up to 4 letters?

  6. anonymous
    • 5 years ago
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    no, the problem is that we want to find the number of groups with four letters from the word abcad for example when we have MATHEMATICS and want to find all the words that can be created without repeating the letters we do: 11!/(2!*2!*2!) because M and A and T are 2 times

  7. anonymous
    • 5 years ago
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    but here I have five letters abcad and want to find only groups with 4 letters

  8. anonymous
    • 5 years ago
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    so I thought it might be 4!/2!

  9. anonymous
    • 5 years ago
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    Ah, but finding groups of letters is different than constructing words.

  10. anonymous
    • 5 years ago
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    So you want to find all the different arrangements of 4 of the letters of abcad such that no letter repeats?

  11. anonymous
    • 5 years ago
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    yes, exactly

  12. anonymous
    • 5 years ago
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    Well there are only 4 unique letters, so you just need to pick where to put each one: \[{4 \choose 1}*{3 \choose 1}*{2\choose 1} = 4!\]

  13. anonymous
    • 5 years ago
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    yes, but as we have the letter a twice, the probability to get a is not the sam as b c or d I am confused, :S

  14. anonymous
    • 5 years ago
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    But the two a's are indistinguishable So: \[ba_1c = ba_2c\] right?

  15. anonymous
    • 5 years ago
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    did you read the example with the word MATHEMATICS I wrote?

  16. anonymous
    • 5 years ago
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    I think it is like that

  17. anonymous
    • 5 years ago
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    Well you reasoned that it would be 11!/(2!*2!*2!). I'm not sure that's correct, but if it is then in this case you'd have 5!/2!

  18. anonymous
    • 5 years ago
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    yea, the MATHEMATICS example is right, I got it from my book and I think is not 5! but 4! because we need groups of four so 4!/2! anyway, I am not sure because I dont have an anser in my book

  19. anonymous
    • 5 years ago
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    No, because the Mathmatics example you cannot compose words with 11 letters having no repeats when you only have 7 letters {m,a,t,h,i,c,s}

  20. anonymous
    • 5 years ago
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    yes we can, thats why we divide by2! three times: if we had all the letters different we would have 11! combinations, but as we have repetitions we have 11!/(2!*2!*2!) combinations, so it makes sense

  21. anonymous
    • 5 years ago
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    Right, so if you have all 5 letters from abcad you'd have 5! combinations.

  22. anonymous
    • 5 years ago
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    yes, but we need 4 and here I get messed up :) anyway, thanx a lot

  23. anonymous
    • 5 years ago
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    hey, btw, I asked 2 other problems with probability can you help me plz

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