How many distinct permutations can be made from the letters of the word “infinity”?
Answer
619.5k+ views
Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, we need to know how to find the factorial for natural numbers. We need to know the formula for finding several possible distinct permutations. We need to compare the given data with the formula of finding several possible distinct words in the given word.
Complete step-by-step answer:
The given word is shown below,
Infinity \[ \to \left( 1 \right)\]
We know that the formula for finding several possible distinct permutations is,
\[\dfrac{{n!}}{{\left( {{r_1}!} \right)\left( {{r_2}!} \right)\left( {{r_3}!} \right)......}} \to \left( 2 \right)\]
Here \[n\] is the total number of types in the given sequence.
\[{r_1}\] are the first type, \[{r_2}\] are the second type, \[{r_3}\] are the third type.
So, we can also write,
\[{r_1} + {r_2} + {r_3} + ..... = n \to \left( 3 \right)\]
In this question, we have the word
Infinity
This word has eight letters. So, we take \[n = 8\] .
Out of eight letters, the letter I am present in three times. So, we take \[{r_1} = 3\] .
Next, we can see the letter n is present two times in the given word. So, we take \[{r_2} = 2\]
Next, we can see the letter f is present one time in the given word. So, we take \[{r_3} = 1\]
Next, we can see that t is present one time in the given word. So, we take \[{r_4} = 1\] .
Next, we can see that y is present one time. So, we take \[{r_5} = 1\] .
By using the values of \[n,{r_1},{r_2},{r_3},{r_4}\] and \[{r_5}\] in the equation \[\left( 2 \right)\] , we get
The required number of permutations
\[ = \dfrac{{n!}}{{\left( {{r_1}!} \right)\left( {{r_2}!} \right)\left( {{r_3}!} \right)\left( {{r_4}!} \right)\left( {{r_5}!} \right)}}\]
\[ = \dfrac{{8!}}{{\left( {3!} \right)\left( {2!} \right)\left( {1!} \right)\left( {1!} \right)\left( {1!} \right)}}\]
\[ = \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8}}{{\left( {1 \times 2 \times 3} \right)\left( {1 \times 2} \right)\left( 1 \right)\left( 1 \right)\left( 1 \right)}}\]
So, we get
Total required number of permutations \[ = \dfrac{{4 \times 5 \times 6 \times 7 \times 8}}{2}\] \[ = 3360\]
So, the final answer is,
Total required number of permutations \[ = 3360\]
So, the correct answer is “ \[ = 3360\] ”.
Note: Note that the value of factorial \[1\] is always one. Also, note that the denominator term would not be equal to zero. Remember the formula for finding the required number of permutations to solve these types of questions. Also, note that if we want to find \[n!\] , we would multiply the following sequence,
\[1 \times 2 \times 3 \times ....n\]
Also, note that the total number of letters in the given word is mentioned as \[n\] . Also, note that we would take the maximum repeated term as \[{r_1}\] .
Complete step-by-step answer:
The given word is shown below,
Infinity \[ \to \left( 1 \right)\]
We know that the formula for finding several possible distinct permutations is,
\[\dfrac{{n!}}{{\left( {{r_1}!} \right)\left( {{r_2}!} \right)\left( {{r_3}!} \right)......}} \to \left( 2 \right)\]
Here \[n\] is the total number of types in the given sequence.
\[{r_1}\] are the first type, \[{r_2}\] are the second type, \[{r_3}\] are the third type.
So, we can also write,
\[{r_1} + {r_2} + {r_3} + ..... = n \to \left( 3 \right)\]
In this question, we have the word
Infinity
This word has eight letters. So, we take \[n = 8\] .
Out of eight letters, the letter I am present in three times. So, we take \[{r_1} = 3\] .
Next, we can see the letter n is present two times in the given word. So, we take \[{r_2} = 2\]
Next, we can see the letter f is present one time in the given word. So, we take \[{r_3} = 1\]
Next, we can see that t is present one time in the given word. So, we take \[{r_4} = 1\] .
Next, we can see that y is present one time. So, we take \[{r_5} = 1\] .
By using the values of \[n,{r_1},{r_2},{r_3},{r_4}\] and \[{r_5}\] in the equation \[\left( 2 \right)\] , we get
The required number of permutations
\[ = \dfrac{{n!}}{{\left( {{r_1}!} \right)\left( {{r_2}!} \right)\left( {{r_3}!} \right)\left( {{r_4}!} \right)\left( {{r_5}!} \right)}}\]
\[ = \dfrac{{8!}}{{\left( {3!} \right)\left( {2!} \right)\left( {1!} \right)\left( {1!} \right)\left( {1!} \right)}}\]
\[ = \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8}}{{\left( {1 \times 2 \times 3} \right)\left( {1 \times 2} \right)\left( 1 \right)\left( 1 \right)\left( 1 \right)}}\]
So, we get
Total required number of permutations \[ = \dfrac{{4 \times 5 \times 6 \times 7 \times 8}}{2}\] \[ = 3360\]
So, the final answer is,
Total required number of permutations \[ = 3360\]
So, the correct answer is “ \[ = 3360\] ”.
Note: Note that the value of factorial \[1\] is always one. Also, note that the denominator term would not be equal to zero. Remember the formula for finding the required number of permutations to solve these types of questions. Also, note that if we want to find \[n!\] , we would multiply the following sequence,
\[1 \times 2 \times 3 \times ....n\]
Also, note that the total number of letters in the given word is mentioned as \[n\] . Also, note that we would take the maximum repeated term as \[{r_1}\] .
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