perms and combs (1 Viewer)

[Vampire]

Hey, i'm having a little trouble with this q:

The letters of the word AUSTRALIA are arranged so two A's must be kept together, while the other A cannot be next to them, in how many ways can this be done.

I get a different answer to other people I've asked. Any help would be appreciated.

 

[ishq]

Okay, so I may be wrong -

There are 8 spots, if we count the 2A's together as 1.

[AA] _ _ _ _ _ _ _

The third A cannot be the immediate spot after [AA]. So, of the 7 letters left, only 6 can take that spot.
After that, the 6 letters left can be arranged in 6! ways.

So, total number of ways would be 6x6!. I'm not sure if we add a 3C2 because all the three A's are the same - so no difference as to which A is chosen to be where.

Am I completely off track?

 

[damo676767]

threat two of the a's as 1 letter

so there are 8 letters and 2 thant can be together

= 8! * (1 - 7/28)

= 30,240

8! is the amount of ways they can be aranged
7 is the amount of pairs that are together in any 1 combo
28 is the amount of pairs of letters

 

[ishq]

threat two of the a's as 1 letter

so there are 8 letters and 2 thant can be together

= 8! * (1 - 7/28)

= 30,240

8! is the amount of ways they can be aranged
7 is the amount of pairs that are together in any 1 combo
28 is the amount of pairs of letters

But you're forgetting that the third A cannot be next to the [AA]! That is included in the above calculation...

 

[damo676767]

But you're forgetting that the third A cannot be next to the [AA]! That is included in the above calculation...

no im not

thats why im timsing it by (1 - 7/28)

that gets rid of all the solution where AA and A are together

i assumed all the A's were identical, if they werent then the solution should be timsed by 6, but i think they are

 

[KFunk]

Your answer agrees with mine. A way I find helpful is to consider the spaces between letters:

₁U₂S₃T₄R₅L₆I₇

There are 6! ways to arrange the non-A letters. There are ⁷C₂ ways to pick two spaces between letters and 2 ways to arrange AA and A between these two spaces giving:

2 x 6! x ⁷C₂ = 30,240 arrangements

 

[Famine]

ishq, I think your answer neglects the fact that the two A's together and the other A can be arranged in different spots, ie

[AA] _ _ _ _ _ _ A
_ _ [AA] _ _ _ _ A

etc, which brings in a LOT more combinations, as damo676767's calculations show.

 

[Abtari]

just one thing to be clarified for myself,

some people have C's and some have factorials in their solution (it always happens with perms and combs)... does it matter which 'approach' you take in solving these types of questions? cos my ans:

6X7X6! = 30240 as well...

 

[Vampire]

I get it now, thx a lot

 

[100percent]

just one thing to be clarified for myself,

some people have C's and some have factorials in their solution (it always happens with perms and combs)... does it matter which 'approach' you take in solving these types of questions? cos my ans:

6X7X6! = 30240 as well...

not really, since
2*7C2=
2*7!/5!*2!=
7x6

 

[ishq]

ishq, I think your answer neglects the fact that the two A's together and the other A can be arranged in different spots, ie

[AA] _ _ _ _ _ _ A
_ _ [AA] _ _ _ _ A

etc, which brings in a LOT more combinations, as damo676767's calculations show.

You're right.
Thanks for that

I had to multiply mine by 7 to account for the 7 positions it can take.
So: 6x6!x7 = 30240

Cheers