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# Permutations

0
104
7
+206

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

Aug 14, 2022

#1
+117852
0

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)

Aug 15, 2022
#2
+206
+1

Answer was incorrect in the app and I also don't understad tying the the two c's together

Aug 15, 2022
#7
+117852
+1

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.

Melody  Aug 16, 2022
edited by Melody  Aug 16, 2022
#3
+2453
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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...

Aug 15, 2022
#6
+2453
0

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.

BuilderBoi  Aug 15, 2022
#4
0

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’?

--. .-

Aug 15, 2022
#5
+2396
+1

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

--. .-

GingerAle  Aug 15, 2022