
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Answer
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Hint: We are given a word ASSASSINATION which consists of 13 letters with repeated terms. We will first pick all the S together and make a single unit and then we will check how many digits are left as we can see N, I and A are repeated and so we use the formula \[\dfrac{n!}{{{P}_{1}}!{{P}_{2}}!{{P}_{3}}!}\] to find the number of words formed.
Complete step-by-step answer:
We are given a word ASSASSINATION which is 13 letter word. First we will check which letter appears how many times. In ASSASSINATION, we see that S comes 4 times, A comes 3 times, I comes 2 times and N comes 2 times.
Now, we are asked to find all those words formed using the letters of the word ASSASSINATION such that all the S’s are together.
So, first of all, we keep all 4 S together and consider it as 1 unit – SSSS.
Now, we can see that we are left with 10 letters, i.e. SSSS AAANNIITO.
Now, we have to make words using these 10 letters. We have to calculate how many letters can be formed using these 10 letters.
As we can see that in the given 10 letters, SSSS AAANNIITO, some letters are repeated. So, for such cases we have the formula,
\[\text{Number of words formed}=\dfrac{n!}{{{P}_{1}}!{{P}_{2}}!{{P}_{3}}!}\]
where n is the total number, \[{{P}_{1}}\] is the number of times the first letter is repeated, \[{{P}_{2}}\] is the number of times the second letter is repeated and so on.
Now, as we know that A is repeated 3 times, so,
\[{{P}_{1}}=3\]
And N is repeated 2 times, so,
\[{{P}_{2}}=2\]
And lastly I is repeated as 2 times, so,
\[{{P}_{3}}=2\]
And the total letters we have is 10. So, n = 10.
Now, putting these in the above formula, we get,
\[\Rightarrow \text{Number of words formed}=\dfrac{10!}{3!2!2!}\]
Now, simplifying we get,
\[\Rightarrow \text{Number of words formed}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3!}{3!2!2!}\]
Cancelling the like terms, we get,
\[\Rightarrow \text{Number of words formed}=10\times 9\times 8\times 7\times 6\times 5\]
\[\Rightarrow \text{Number of words formed}=151200\]
So, we have that number of words where all the S’s are together are 151200.
Note: As the letters are being repeated, so we cannot use n! to find the total word formed. We have to cancel the possible repeated words. So, we use the formula,
\[\text{Number of words formed}=\dfrac{n!}{{{P}_{1}}!{{P}_{2}}!{{P}_{3}}!......{{P}_{n}}!}\]
When we put all the S together and make a single unit, it will add up with the rest of the letter. While making a new word, we can shift its location and make new words, it is contributing to the words. So, we have a total 9 + 1 = 10 letters.
Complete step-by-step answer:
We are given a word ASSASSINATION which is 13 letter word. First we will check which letter appears how many times. In ASSASSINATION, we see that S comes 4 times, A comes 3 times, I comes 2 times and N comes 2 times.
Now, we are asked to find all those words formed using the letters of the word ASSASSINATION such that all the S’s are together.
So, first of all, we keep all 4 S together and consider it as 1 unit – SSSS.
Now, we can see that we are left with 10 letters, i.e. SSSS AAANNIITO.
Now, we have to make words using these 10 letters. We have to calculate how many letters can be formed using these 10 letters.
As we can see that in the given 10 letters, SSSS AAANNIITO, some letters are repeated. So, for such cases we have the formula,
\[\text{Number of words formed}=\dfrac{n!}{{{P}_{1}}!{{P}_{2}}!{{P}_{3}}!}\]
where n is the total number, \[{{P}_{1}}\] is the number of times the first letter is repeated, \[{{P}_{2}}\] is the number of times the second letter is repeated and so on.
Now, as we know that A is repeated 3 times, so,
\[{{P}_{1}}=3\]
And N is repeated 2 times, so,
\[{{P}_{2}}=2\]
And lastly I is repeated as 2 times, so,
\[{{P}_{3}}=2\]
And the total letters we have is 10. So, n = 10.
Now, putting these in the above formula, we get,
\[\Rightarrow \text{Number of words formed}=\dfrac{10!}{3!2!2!}\]
Now, simplifying we get,
\[\Rightarrow \text{Number of words formed}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3!}{3!2!2!}\]
Cancelling the like terms, we get,
\[\Rightarrow \text{Number of words formed}=10\times 9\times 8\times 7\times 6\times 5\]
\[\Rightarrow \text{Number of words formed}=151200\]
So, we have that number of words where all the S’s are together are 151200.
Note: As the letters are being repeated, so we cannot use n! to find the total word formed. We have to cancel the possible repeated words. So, we use the formula,
\[\text{Number of words formed}=\dfrac{n!}{{{P}_{1}}!{{P}_{2}}!{{P}_{3}}!......{{P}_{n}}!}\]
When we put all the S together and make a single unit, it will add up with the rest of the letter. While making a new word, we can shift its location and make new words, it is contributing to the words. So, we have a total 9 + 1 = 10 letters.
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