anonymous
  • anonymous
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 !
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[{5\choose 1} * {4 \choose 1}*{3 \choose 1}*{2 \choose 1} = 5!\]
anonymous
  • anonymous
I do not understand? as the letter a is two times we should divide with 2! is it right?
anonymous
  • anonymous
Ack! I didn't notice a was in there twice.

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anonymous
  • anonymous
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.
anonymous
  • anonymous
Or did you mean groups with up to 4 letters?
anonymous
  • anonymous
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
anonymous
  • anonymous
but here I have five letters abcad and want to find only groups with 4 letters
anonymous
  • anonymous
so I thought it might be 4!/2!
anonymous
  • anonymous
Ah, but finding groups of letters is different than constructing words.
anonymous
  • anonymous
So you want to find all the different arrangements of 4 of the letters of abcad such that no letter repeats?
anonymous
  • anonymous
yes, exactly
anonymous
  • anonymous
Well there are only 4 unique letters, so you just need to pick where to put each one: \[{4 \choose 1}*{3 \choose 1}*{2\choose 1} = 4!\]
anonymous
  • anonymous
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
anonymous
  • anonymous
But the two a's are indistinguishable So: \[ba_1c = ba_2c\] right?
anonymous
  • anonymous
did you read the example with the word MATHEMATICS I wrote?
anonymous
  • anonymous
I think it is like that
anonymous
  • anonymous
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!
anonymous
  • anonymous
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
anonymous
  • anonymous
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}
anonymous
  • anonymous
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
anonymous
  • anonymous
Right, so if you have all 5 letters from abcad you'd have 5! combinations.
anonymous
  • anonymous
yes, but we need 4 and here I get messed up :) anyway, thanx a lot
anonymous
  • anonymous
hey, btw, I asked 2 other problems with probability can you help me plz

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