Fun harder counting problem (1 Viewer)

[king.rafa]

how many arrangements of australia if no 2 a's can be together.

 

[lychnobity]

how many arrangements of australia if no 2 a's can be together.

Answer = (6! x 7C3)/3

Explanation: There are 6! ways to arrange the other letters (u,s,t,r,l,i). There are 7 "spaces" (in between the 6 letters, including the space at the end of the line). Of these 7 spaces, you are going to fill up 3 of them. Therefore you do 7C3

 

[Timothy.Siu]

Answer = (6! x 7C3)/3

Explanation: There are 6! ways to arrange the other letters (u,s,t,r,l,i). There are 7 "spaces" (in between the 6 letters, including the space at the end of the line). Of these 7 spaces, you are going to fill up 3 of them. Therefore you do 7C3

why do u divide by 3?
i'm not too good at these but i think it should just be 6!x7C3

 

[lychnobity]

why do u divide by 3?
i'm not too good at these but i think it should just be 6!x7C3

I was thinking that you have to divide by 3 because the As are the same. It would lessen the no. ways seeing as the As swapping places wouldn't form a new word.

Say, the letters B, C and D took the place of the 3 As (given same conditions as the question above). B & C swapping places anywhere would form a new word, and so would swapping C & D. But since the As are the same you'd have to divide by 3 to account for it.

 

[Drongoski]

I was thinking that you have to divide by 3 because the As are the same. It would lessen the no. ways seeing as the As swapping places wouldn't form a new word.

Say, the letters B, C and D took the place of the 3 As (given same conditions as the question above). B & C swapping places anywhere would form a new word, and so would swapping C & D. But since the As are the same you'd have to divide by 3 to account for it.

After reading this I suspect you are right. At 1st I thought Tim was right but realise I forgot the 3 A's are indistinguishable.

Edit

Correction: shouldn't that be dividing by '3!' rather than '3' in that case ??

 

[Timothy.Siu]

but ur using C not P

if u did 7P3 then u wud have to divide
but since you're using C, those repeating cases aren't counted

 

[Drongoski]

but ur using C not P

if u did 7P3 then u wud have to divide
but since you're using C, those repeating cases aren't counted

I didn't read carefully; should look before I leap. Thanks.

 

[Timothy.Siu]

I didn't read carefully; should look before I leap. Thanks.

lol i was talking to the other dude, haha

 

[king.rafa]

would the answer be 9!/3! - ( 7! + 7!*6)

 

[Timothy.Siu]

would the answer be 9!/3! - ( 7! + 7!*6)

yeah thats the answer, but how did u get 7!*6

 

[gurmies]

No need to consider the repeating A's as Tim poined out...