Question

The number of arrangements of the letters \[abcd\] in which neither \[a,b\] nor \[c,d\] come together is:
A) 6
B) 12
C) 16
D) None

Answer 1

Hint:
Here, we will first find the total number of ways of the arrangement of the given letters. Then we will find the possible number of ways that both the letters come together. After which we will subtract it from the total number of ways to get the required answer.

Complete step by step solution:
We have to find the total number of arrangements of the letters \[abcd\].
Total number of arrangements of the letters \[abcd = 4!\]
Computing the factorial, we get
\[ \Rightarrow \] Total number of arrangements of the letters \[abcd = 4 \times 3 \times 2 \times 1\]
\[ \Rightarrow \] Total number of arrangements of the letters \[abcd = 24\]
Now, we have to find the number of arrangements in which only \[a,b\] come together.
Since \[a\] and \[b\] come together, we consider \[a\] and \[b\] as one single identity and thus can be arranged in 3 ways. Also, individually, \[a\] and \[b\] can be arranged in 2 ways. So, we get
Number of arrangements\[ = 3!2!\]
Computing the factorial, we get
\[ \Rightarrow \] Number of arrangements\[ = 3 \times 2 \times 1 \times 2 \times 1\]
\[ \Rightarrow \] Number of arrangements\[ = 12\]
Now, we have to find the number of arrangements in which only \[c,d\]come together.
Since \[c\] and \[d\] come together, we will consider \[c\] and \[d\] as one single identity and thus can be arranged in 3 ways. Also, individually, \[c\] and \[d\] can be arranged in 2 ways. So, we get
Number of arrangements \[ = 3!2!\]
Computing the factorial, we get
\[ \Rightarrow \] Number of arrangements\[ = 3 \times 2 \times 1 \times 2 \times 1\]
\[ \Rightarrow \] Number of arrangements\[ = 12\]
Now, we have to find the number of arrangements in which both \[a\] and \[b\], as well as \[c\] and \[d\], come together.
Since \[a\] and \[b\]as well as \[c\] and \[d\] come together it can be arranged in 2 ways and both these letters can be arranged in 2 ways separately. So, we get
Number of arrangements\[ = 2!2!2!\]
Computing the factorial, we get
\[ \Rightarrow \] Number of arrangements\[ = 2 \times 1 \times 2 \times 1 \times 2 \times 1\]
\[ \Rightarrow \] Number of arrangements\[ = 8\]
Now, we have to find the total number of arrangements by subtracting the number of ways in all these case from the total number of ways, thus we get
Number of required arrangements \[ = 24 - 12 - 12 + 8\]
Adding and subtracting the terms, we get
\[ \Rightarrow \] Number of required arrangements \[ = 8\]

Therefore, the number of arrangements of the letters \[abcd\] in which neither \[a,b\] nor \[c,d\] come together is none.

Note:
We have used the concept of permutation. A permutation is defined as the arrangement of letters, numbers, or some elements in a set. It gives us the number of ways that the elements in a set are arranged. Combination is defined as the selection of objects. Both are similar but in permutations, the order is important while in combinations order is not important. Factorial is defined as the numbers multiplied in descending order till unity.