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

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- anonymous

\[{5\choose 1} * {4 \choose 1}*{3 \choose 1}*{2 \choose 1} = 5!\]

- anonymous

I do not understand?
as the letter a is two times we should divide with 2!
is it right?

- anonymous

Ack! I didn't notice a was in there twice.

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## More answers

- 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

Or did you mean groups with up to 4 letters?

- 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

but here I have five letters abcad and want to find only groups with 4 letters

- anonymous

so I thought it might be 4!/2!

- anonymous

Ah, but finding groups of letters is different than constructing words.

- anonymous

So you want to find all the different arrangements of 4 of the letters of abcad such that no letter repeats?

- anonymous

yes, exactly

- 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

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

But the two a's are indistinguishable
So:
\[ba_1c = ba_2c\]
right?

- anonymous

did you read the example with the word MATHEMATICS I wrote?

- anonymous

I think it is like that

- 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

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

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

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

Right, so if you have all 5 letters from abcad you'd have 5! combinations.

- anonymous

yes, but we need 4 and here I get messed up
:)
anyway, thanx a lot

- anonymous

hey, btw, I asked 2 other problems with probability
can you help me plz

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