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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 !
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{5\choose 1} * {4 \choose 1}*{3 \choose 1}*{2 \choose 1} = 5!\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I do not understand? as the letter a is two times we should divide with 2! is it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ack! I didn't notice a was in there twice.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0Or did you mean groups with up to 4 letters?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, 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
 5 years ago
Best ResponseYou've already chosen the best response.0but here I have five letters abcad and want to find only groups with 4 letters

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I thought it might be 4!/2!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, but finding groups of letters is different than constructing words.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you want to find all the different arrangements of 4 of the letters of abcad such that no letter repeats?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0But the two a's are indistinguishable So: \[ba_1c = ba_2c\] right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you read the example with the word MATHEMATICS I wrote?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0yea, 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
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0Right, so if you have all 5 letters from abcad you'd have 5! combinations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but we need 4 and here I get messed up :) anyway, thanx a lot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey, btw, I asked 2 other problems with probability can you help me plz
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