
In how many different ways can you arrange the letters of the word INTERNET?
Answer
545.4k+ views
Hint: In the above-given problem first we are going to count the number of letters present in the word INTERNET. By using permutation, we can say that it can be arranged in the same number of factorials as the number of letters present in the word. Then we have to observe the number of such letters present in the word INTERNET occurring two or more than two times. Because this will give the same arrangement, not the distinct one. Therefore, we have to divide with the number of letters occurring two or more than two times to obtain the exact number of distinct arrangements.
Complete step by step solution:
Counting the number of letters present in the word given in the problem and writing it as following,
Number of letters present in Internet \[ = 8\]
By using permutation total number of arrangements can be written as following,
Total number of arrangements \[ = 8!\]
Now counting the number of letters occurring two or more than two times and writing as following,
‘N’ is occurring two times,
‘E’ is occurring two times,
‘T’ is occurring two times,
Therefore, two ‘N’ can be arranged \[ = 2!\]
Also, two ‘N’ can be arranged in \[ = 2!\]
Also, two ‘N’ can be arranged in \[ = 2!\]
Dividing with the above obtained arrangements and writing as follows,
\[ = \dfrac{{8!}}{{2!2!2!}}\]
Simplifying and writing it as following,
\[ = \dfrac{{8!}}{{2! \times 2! \times 2!}}\]
Expanding factorial, we get
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1 \times 2 \times 1}}\]
Multiplying denominator, we get
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{8}\]
Cancelling numerator with the denominator, we get
\[ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
Simplifying by multiplying, we get
\[ = 5040\]
Hence, we got \[5040\] total number of distinct arrangements.
Note:
A permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of arrangements matter.
Complete step by step solution:
Counting the number of letters present in the word given in the problem and writing it as following,
Number of letters present in Internet \[ = 8\]
By using permutation total number of arrangements can be written as following,
Total number of arrangements \[ = 8!\]
Now counting the number of letters occurring two or more than two times and writing as following,
‘N’ is occurring two times,
‘E’ is occurring two times,
‘T’ is occurring two times,
Therefore, two ‘N’ can be arranged \[ = 2!\]
Also, two ‘N’ can be arranged in \[ = 2!\]
Also, two ‘N’ can be arranged in \[ = 2!\]
Dividing with the above obtained arrangements and writing as follows,
\[ = \dfrac{{8!}}{{2!2!2!}}\]
Simplifying and writing it as following,
\[ = \dfrac{{8!}}{{2! \times 2! \times 2!}}\]
Expanding factorial, we get
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1 \times 2 \times 1}}\]
Multiplying denominator, we get
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{8}\]
Cancelling numerator with the denominator, we get
\[ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
Simplifying by multiplying, we get
\[ = 5040\]
Hence, we got \[5040\] total number of distinct arrangements.
Note:
A permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of arrangements matter.
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