Question

The number of arrangements of the letters ‘a b c d’ in which neither a, b nor c, d come together is:
A. 6
B. 12
C. 16
D. None

Answer 1

Hint: First we will find the total possible arrangement of four given letters and considering three cases, where case 1 has a and b as 1, case 2 has c and d as 1 and case 3 has a and b and c and d both come together using the formula to calculate factorials is \[n! = n \times \left( {n - 1} \right)!\], where \[n\] is the number of items in the above expression. Then we will find the number of ways when neither a b nor c d comes together by subtracting the above case from the total number of arrangements.

Complete step by step answer:

We are given the letters ‘a b c d’.

Finding the total possible arrangement of four given letters, we get
\[
   \Rightarrow 4! \\
   \Rightarrow 4 \times 3 \times 2 \times 1 \\
   \Rightarrow 24 \\
 \]
So, while selecting the letters for arrangement we will consider all the cases.
Case 1:
 Take a, b as 1 unit
In this case, we will arrange a, b as 1 unit and this can be arranged in \[2!\] ways.
Since we have three choices to find the number of ways so that we have to select two letters a and b, we get
\[ \Rightarrow 3! \times 2\]
Using the formula to calculate factorials is \[n! = n \times \left( {n - 1} \right)!\], where \[n\] is the number of items in the above expression, we get

\[
   \Rightarrow 3 \times 2 \times 1 \times 2 \\
   \Rightarrow 12{\text{ ways}} \\
 \]
Thus, there are 12 ways.
Case 2:
 Take c, d as 1 unit
In this case, we will arrange c, d as 1 unit and this can be arranged in \[2!\] ways
Since we have three choices to find the number of ways so that we have to select two letters c and d, we get
\[ \Rightarrow 3! \times 2\]
Using the formula to calculate factorials is \[n! = n \times \left( {n - 1} \right)!\], where \[n\] is the number of items in the above expression, we get

\[
   \Rightarrow 3 \times 2 \times 1 \times 2 \\
   \Rightarrow 12{\text{ ways}} \\
 \]
Thus, there are 12 ways.

Case 3:
 Take a, b and c, d both come together
In this case, we will arrange a, b as 1 unit, which is arranged in \[2!\] ways and c, d as 1 unit, which are arranged in \[2!\] ways.
Since we have two choices to find the number of ways so that we have to select two letters a and b and two letters c and d, we get
\[ \Rightarrow 2! \times 2 \times 2\]
Using the formula to calculate factorials is \[n! = n \times \left( {n - 1} \right)!\], where \[n\] is the number of items in the above expression, we get

\[
   \Rightarrow 2 \times 1 \times 2 \times 2 \\
   \Rightarrow 8{\text{ ways}} \\
 \]
Thus, there are 8 ways.

Now, we will find the number of ways when neither a b nor c d comes together by subtracting the case 1, case 2 from the total number of arrangements and add case 3 in the obtained expression, we get
\[
   \Rightarrow 24 - 12 - 12 + 8 \\
   \Rightarrow 8{\text{ ways}} \\
 \]
Since none of the options is 8, thus options A, B and C are correct.
Thus, option D is correct.

Note: In solving these types of questions, generally students get confused between factorials, combination, and permutation. If you have to use combination \[{}^n{C_r}\] and if you have to arrange use permutation \[{}^n{P_r}\], it is very tricky to use. Some students forget to consider all possibilities or else might get the wrong answer, because if some case is missed then it will result in the wrong answer. We should take care of substitution in the final equation to find the required answer.