
The number of arrangements of the letters a b c d in which neither a, b nor c, d come together.
A) 6
B) 12
C) 16
D) None of these
Answer
443.5k+ views
Hint: In this question first we find the total number of ways to arrange 4 letters after that we find the number of possible ways in which \[{\text{a & b}}\] and \[{\text{c & d}}\] together. After that, we subtract it from the total number of ways and we will get the required answer.
Complete step by step solution: As we all know letters a b c d can be arranged in \[4!\]
\[4! = 4 \times 3 \times 2 \times 1\]
\[ \Rightarrow 4! = 24\]
\[ \Rightarrow \] Total possible arrangement of letters a b c d is 24.
In order to calculate neither \[{\text{a & b}}\] nor \[{\text{c & d}}\] come together first we calculate the cases when \[{\text{a & b}}\] comes together and \[{\text{c & d}}\] together.
So, take \[{\text{a & b}}\] as one entity so we have total 3 letters.
\[ \Rightarrow \] 3 letters can be arranged in \[3!\] and \[{\text{a & b}}\] itself arranged in \[2!\]
\[ \Rightarrow \] Case when \[{\text{a & b}}\] comes together then letters can be arranged in \[3! \times 2!\]\[ = 3 \times 2 \times 1 \times 2 \times 1 = 12\] ways.
Similarly,
Take \[{\text{c & d}}\] as one entity so we have total 3 letters.
\[ \Rightarrow \] 3 letters can be arranged in \[3!\] and \[{\text{c & d}}\] itself arranged in \[2!\]
\[ \Rightarrow \] Case when \[{\text{c & d}}\] comes together then letters can be arranged in \[3! \times 2!\]\[ = 3 \times 2 \times 1 \times 2 \times 1 = 12\] ways.
Now, the case when \[{\text{a & b}}\] and \[{\text{c & d}}\] both come together.
\[ \Rightarrow \] Case when \[{\text{a & b}}\] and \[{\text{c & d}}\] both come together then letters can be arranged in \[2! \times 2! \times 2! = 2 \times 2 \times 2 = 8\] ways.
So, the total number of arrangement when neither \[{\text{a & b}}\] nor \[{\text{c & d}}\] come together \[ = 24 - 12 - 12 + 8\]
\[ = 8\]
8 number of ways the letters a b c d can arrange in which neither a,b nor c,d come together.
Hence, option D. None of these is the correct answer.
Note: Here we have used the concept of permutation. Permutation can be defined as arranging or rearranging the elements of a set. It gives the number of ways a certain number of objects can be arranged. Combinations can be defined as selecting a number of elements from a given set. Both permutations and combinations are similar except that in permutations, the order is important and in combinations the order is not important.
Complete step by step solution: As we all know letters a b c d can be arranged in \[4!\]
\[4! = 4 \times 3 \times 2 \times 1\]
\[ \Rightarrow 4! = 24\]
\[ \Rightarrow \] Total possible arrangement of letters a b c d is 24.
In order to calculate neither \[{\text{a & b}}\] nor \[{\text{c & d}}\] come together first we calculate the cases when \[{\text{a & b}}\] comes together and \[{\text{c & d}}\] together.
So, take \[{\text{a & b}}\] as one entity so we have total 3 letters.
\[ \Rightarrow \] 3 letters can be arranged in \[3!\] and \[{\text{a & b}}\] itself arranged in \[2!\]
\[ \Rightarrow \] Case when \[{\text{a & b}}\] comes together then letters can be arranged in \[3! \times 2!\]\[ = 3 \times 2 \times 1 \times 2 \times 1 = 12\] ways.
Similarly,
Take \[{\text{c & d}}\] as one entity so we have total 3 letters.
\[ \Rightarrow \] 3 letters can be arranged in \[3!\] and \[{\text{c & d}}\] itself arranged in \[2!\]
\[ \Rightarrow \] Case when \[{\text{c & d}}\] comes together then letters can be arranged in \[3! \times 2!\]\[ = 3 \times 2 \times 1 \times 2 \times 1 = 12\] ways.
Now, the case when \[{\text{a & b}}\] and \[{\text{c & d}}\] both come together.
\[ \Rightarrow \] Case when \[{\text{a & b}}\] and \[{\text{c & d}}\] both come together then letters can be arranged in \[2! \times 2! \times 2! = 2 \times 2 \times 2 = 8\] ways.
So, the total number of arrangement when neither \[{\text{a & b}}\] nor \[{\text{c & d}}\] come together \[ = 24 - 12 - 12 + 8\]
\[ = 8\]
8 number of ways the letters a b c d can arrange in which neither a,b nor c,d come together.
Hence, option D. None of these is the correct answer.
Note: Here we have used the concept of permutation. Permutation can be defined as arranging or rearranging the elements of a set. It gives the number of ways a certain number of objects can be arranged. Combinations can be defined as selecting a number of elements from a given set. Both permutations and combinations are similar except that in permutations, the order is important and in combinations the order is not important.
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