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Question

Question: The maximum number of different permutation of 4 letters of the word EARTHQUAKE is: A. \(2910\) ...

The maximum number of different permutation of 4 letters of the word EARTHQUAKE is:
A. 29102910
B. 25502550
C. 21902190
D. 20912091

Explanation

Solution

In the given question we will apply three cases of permutation, first is when all four letters are different, second is when two letters are the same and two letters are different, and third is when two letters are the same and other two letters are also same. We will find the values in each case, and then add all three values to get the maximum number of different permutation. Thus we will get the answer.

Complete step by step answer:

We will make all three possible cases for the given question:
Case 1: when all four letters are different:
As we know:
Total Number of letters are=8 = 8 in the given word EARTHQUAKE
Four-word letters are formed by 88 different letters:
8C4×4! 8×7×6×54×3×2×1×4×3×2×1 8×7×6×5 1680  { \Rightarrow ^8}{C_4} \times 4! \\\ \Rightarrow \dfrac{{8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1}} \times 4 \times 3 \times 2 \times 1 \\\ \Rightarrow 8 \times 7 \times 6 \times 5 \\\ \Rightarrow 1680 \\\
Thus we get the number of ways when all four letters are different.
Case 2: when two letters are the same and two letters are different:
2C1×7C2×4!2! 504  \Rightarrow \dfrac{{^2{C_1}{ \times ^7}{C_2} \times 4!}}{{2!}} \\\ \Rightarrow 504 \\\
Thus we get the number of ways when two letters are the same and two letters are different.
Case 3: is when two letters are the same and the other two letters are also the same:
Four-word letters are formed by this case=4!2!2! = \dfrac{{4!}}{{2!2!}}
6\Rightarrow 6
Thus we get the number of ways when two letters are the same and the other two letters are also the same.
Now add all the total number of ways.
Thus we get
=1680+504+6 2190  = 1680 + 504 + 6 \\\ \Rightarrow 2190 \\\
So the maximum number of different permutation of 4 letters of the word EARTHQUAKE is21902190.
And this is our answer.
Hence the correct answer is option C.

Note: In the given question we have to remember to make all possible cases according to the question, with doing this we cannot find the total number of ways. We have to make permutation combinations for all possible cases. Then we have to add the number of all possible cases for the total number of ways. Thus we get the correct answer.