
How many different arrangements can be made using all of the letters in the word REARRANGE?
Answer
536.1k+ views
Hint: To solve the above question we will permutation formula. In mathematics, a permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangement matters or choosing only several items from a set of items with a certain order. The general permutation formula is expressed in the following way:
$\Rightarrow P\left( n,k \right)=\dfrac{n!}{\left( n-k \right)!}$, where $n$ is the total number of elements in a set, $k$ is the number of selected elements arranged in a specific order and ! is factorial. The factorial is the product of all positive integers less than or equal to the number preceding the factorial sign.
Complete step by step answer:
Now in the given question we have to arrange the word REARRANGE. So, if there are $n$ numbers of elements inside the set then the permutation formula for rearranging is${{P}_{n}}=n!$, and if the words are repeated in the given word then the permutation formula for rearranging is${{P}_{n}}=\dfrac{n!}{r!s!......}$ , where $r,s$ the words which are repeating in the given word.
The given word is REARRANGE. In this word the total number of elements are$9$ and the repeating elements are $3R’s,2A’s$ and $2E’s$.
Now we will put all above values in the formula of the permutation which is:
$\Rightarrow {{P}_{n}}=\dfrac{n!}{r!s!......}$
Now putting values we get,
$\Rightarrow {{P}_{n}}=\dfrac{9!}{3!\times 2!\times 2!}$
Now when we calculate the value of $9!$ from calculator then we get $9!=362,880$and similarly the $3!=6,2!=2$ now putting these values in the above formula we get,
$\begin{align}
& \Rightarrow {{P}_{n}}=\dfrac{362,880}{6\times 2\times 2} \\
& \Rightarrow {{P}_{n}}=\dfrac{362880}{24} \\
& \Rightarrow {{P}_{n}}=15120 \\
\end{align}$
Hence there are $15,120$ ways to write the word REARRANGE.
Note: To solve these types of questions we always have to use the permutation concept. Permutations are frequently confused with another mathematical technique called combination. In combinations, the order of the elements does not matter. But in permutation the order of the elements matters.
$\Rightarrow P\left( n,k \right)=\dfrac{n!}{\left( n-k \right)!}$, where $n$ is the total number of elements in a set, $k$ is the number of selected elements arranged in a specific order and ! is factorial. The factorial is the product of all positive integers less than or equal to the number preceding the factorial sign.
Complete step by step answer:
Now in the given question we have to arrange the word REARRANGE. So, if there are $n$ numbers of elements inside the set then the permutation formula for rearranging is${{P}_{n}}=n!$, and if the words are repeated in the given word then the permutation formula for rearranging is${{P}_{n}}=\dfrac{n!}{r!s!......}$ , where $r,s$ the words which are repeating in the given word.
The given word is REARRANGE. In this word the total number of elements are$9$ and the repeating elements are $3R’s,2A’s$ and $2E’s$.
Now we will put all above values in the formula of the permutation which is:
$\Rightarrow {{P}_{n}}=\dfrac{n!}{r!s!......}$
Now putting values we get,
$\Rightarrow {{P}_{n}}=\dfrac{9!}{3!\times 2!\times 2!}$
Now when we calculate the value of $9!$ from calculator then we get $9!=362,880$and similarly the $3!=6,2!=2$ now putting these values in the above formula we get,
$\begin{align}
& \Rightarrow {{P}_{n}}=\dfrac{362,880}{6\times 2\times 2} \\
& \Rightarrow {{P}_{n}}=\dfrac{362880}{24} \\
& \Rightarrow {{P}_{n}}=15120 \\
\end{align}$
Hence there are $15,120$ ways to write the word REARRANGE.
Note: To solve these types of questions we always have to use the permutation concept. Permutations are frequently confused with another mathematical technique called combination. In combinations, the order of the elements does not matter. But in permutation the order of the elements matters.
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