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I am trying to solve the following question and my logic seems correct but my answer is not matching.

"In how many ways can letters of the word $\text{CALENDAR}$ be arranged so $\text{C}$ and $\text{A}$ are together as well as $\text{N}$ and $\text{D}$ are together?"

Question: What is wrong or missing in my logic explained below?

My Logic

We start with $8$ empty spots along a straight line and fill them by performing the three actions below one after the other.

Action 1 - First Place $\text{C}$ and $\text{A}$ together : $7 \times 2! = 14$.

Action 2 - Then Place $\text{N}$ and $\text{D}$ together after $\text{C}$ and $\text{D}$ have been placed : $5 \times 2! = 10$.

Action 3 - Finally, Place other $4$ different characters in remaining $4$ spots : $4!$

So, total number of ways using Product Rule is $14 \times 10 \times 4! = 3360$

There is something wrong with my logic since the book says $2880$ is the answer.

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    You are double counting arrangements where you first put $CA $ and then the second $A $ to the left, or you first put $AC $ and then the second $A $ to the right. You should be careful as these are being counted twice but represent the same configuration – 2017-01-21
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    Ok. I got it. So in my logic I was counting the same arrangement more than once. – 2017-01-21

2 Answers 2

1

You have the word CALENDAR.
Condition in the problem says C and A are grouped together, let it be X=CA. Again, N and D are grouped together, so let it be Y=ND. Now the modified word is XLEYAR.
Now, since there are 6 letters in modified word, therefore, number of permutations will be,
6!=720
If you notice, the words in X and Y, that is, C and A in X can be arranged in two ways, CA or AC, therefore, number of permutation in which C and A can be arranged is 2!=2.
Similar argument holds for words assigned to Y. We again get, number of ways in which N and D can be arranged = 2! =2.
Now use fundamental principle of counting, remembering the fact that events as defined above are independent, this will give,
Number of ways CALENDAR can be arranged = 2 * 2 * 720 =2880 (ans)

1

Group CA and ND.

So we have 2 groups and 4 letters.

Numbers of ways = 6!

But letters further arranged in 2 ways in each group.

So we have $6! * 2 * 2$

= 720 * 4 = 2880.

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    Why did you say 2*6! for arranging 2 groups and 4 letters? It should be 6! for arranging 6 different objects. – 2017-01-21
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    As we have 2 A's. So multiply the pair with 2. – 2017-01-21
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    Why would you arrange 2 As? There's 2. As can only be arranged in 1 way and not 2 ways. – 2017-01-21
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    Ok I edited my post as before. Because if we includes case of 2 A's then that is more confusing. – 2017-01-21