+275 In how many ways can the letters of the word ACCURACY be rearranged so that no two C's appear as consecutive letters in the rearrangement?
The word accuracy can be arranged 3360 times
+118673 Accuracy
8 letters, 3Csand 2As 8!/(3!*2!) = 3360 As the question said
aCuracy (I tied 2 Cs together)
How many ways can this be arranged
7 letters, 2As
7!/ (2!) = 2520, but if the little c is on either side of the C then it will be the same so I need to divide by 2
2520/2=1260
So the mumber of ways no two cs are together is 3360-1260 = 2100 ways
If you think this is incorrect then please promote a discussion.
Or at least say so and give reason (eg answer is rejected in the app)
+275 Answer was incorrect in the app and I also don't understad tying the the two c's together
+118673 Thanks Jasonkiln,
You are right of course, I cannot just divide by 2 at the end
.I can tie my two Cs together though. becasue any answer with 2 Cs tied together will not be included and I will have to subtract.
So I have
aauryC and I will add the other c later
6!/2! = 360
Now where can the last c go
|a|a|u|r|y|C
It can go where any of those bars are. It can't go after the C becasue that would be the same as before the C Cc = cC
There are 6 bars.
360*6=2160
There are 2160 permutations where 2 or more Cs are together
3360-2160 = 1200
There are 1200 permutations where no 2 Cs are together.
This is now the same as Ginger's answer, Builderboi agrees too. I think his method is similar to mine.
Ginger's method is a little more simple.
+2668 Here's my attempt:
There are 2 cases to consider: exactly 2 C's are next to each other, or 3 C's are next to each other.
If 3 C's are next to each other, there are 6 ways to put them. Then, for the remaining 5 digits, there are \({5! \over 2!} = 60\) ways to order them. So, there are \(6 \times 60 = 360\) ways to order this.
If 2 C's are next to each other, then there are 2 subcases: The 2 C's are on the sides (1 and 2 or 7 and 8) or they are in the center.
If the C's are on the sides, there are 2 ways to choose to put the C's. Then, the third C has 5 choices (can't be right next to the other C's). Then, there are \({5! \over 2!} = 60\) ways to order the remaining digits. So, there are \(2 \times 5 \times 60 = 600\) ways for this case.
Now, if the C's are in the center, there are 5 ways to put the C's. Then, the third C has 4 choices (can't be on either side). Then, there are \({5! \over 2!} = 60\) ways for the remaining digits. So, there are \(5 \times 4 \times 60 = 1200\) ways for this.
So, there are \(3360 - 300-600-1200 = \color{brown}\boxed{1260}\) ways. I think...
+2668 Wait, I see my mistake... I subtracted 300 instead of 360.
So the correct answer becomes \(3360- 360-600-1200 = \color{brown}\boxed{1200}\).
Now, GingerAle and I are in agreement. 
In how many ways can the letters of the word ACCURACY be rearranged so that no two C's appear as consecutive letters in the rearrangement?
The word accuracy can be arranged 3360 times
Restatement of question:
How many arrangements of the letters in the word ACCURACY are there, such that no ‘C’s are adjacent to another ‘C’?
--. .-
+2489 Solution: \(\dfrac{5!}{2!} * \dbinom {6}{3}\)
Description of process:
Removing the ‘C’s from the set gives (5!/2! =60) distinguishable arraignments for the remaining five (5) letters. The three ‘C’s can then occupy three (3) of the six (6) positions between the letters and at either end of the distinguishable arraignments
Graphic demonstrating positions: _A_U_R_A_Y_
Select the three (3) positions as \(\dbinom {6}{3} = 20\)
Total number of arrangements where no ‘C’s are adjacent to another ‘C’ \(= (60 * 20) = 1200 \)
GA
--. .-
