Question

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 !

Answer 1

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

Answer 2

I do not understand?

as the letter a is two times we should divide with 2!
is it right?

Answer 3

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

Answer 4

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.

Answer 5

Or did you mean groups with up to 4 letters?

Answer 6

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

Answer 7

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

Answer 8

so I thought it might be 4!/2!

Answer 9

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

Answer 10

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

Answer 11

yes, exactly

Answer 12

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

Answer 13

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

Answer 14

But the two a's are indistinguishable

So:
\[ba_1c = ba_2c\]
right?

Answer 15

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

Answer 16

I think it is like that

Answer 17

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!

Answer 18

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

Answer 19

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}

Answer 20

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

Answer 21

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

Answer 22

yes, but we need 4 and here I get messed up
:)

anyway, thanx a lot

Answer 23

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