How many unique ways are there to arrange the letters in the word “PRETTY” ?
Answer
563.1k+ views
Hint: We should make use of the concepts of permutations and combinations .We can clearly see that this word has repeated letters. So , we initially need to count the number of letters that consist of the word. And later, we count the number of letters that are repeated and the number of times they are repeated. And then we just have to divide the total number of ways in which we can write the word divided by the factorial of the number of times a letter is repeated.
Complete step by step solution:
We have to divide the total number of ways in which we can write the word by the factorial of the number of times a letter is repeated.
For example we have a few letters together . Let is aabc. The combinations that can be formed are the following :
$1)$ abc
$2)$ acb
$3)$ bac
$4)$ bca
$5)$ cab
$6)$ cba
We can’t really tell whether the first a or the second a is being in these combinations. Just because we don’t know which a is in these combinations, we do not write a combination with first a and the same combinations with the second a since both of them will be the same.
In the same, repeated letter will make up the same word no matter which repetition makes up the word.
After counting we can safely say that the word is made up, $6$letters.
Let us find the number of repeated letters and the number of repetition of each letter.
We can infer from the word that the letter T is repeated for $2$times.
The non-repeating letters are P,R,E,Y.
The total number of ways to write the word “PRETTY” would be $6!$ .
Let us divide it by $2!$ so as to subtract or delete all the same words which are made up of the repetition of the repeated letter.
Upon doing so, we get the following :
$\Rightarrow \dfrac{6!}{2!}=360$
$\therefore $ The number of unique ways there are to arrange the letters in the word “PRETTY” are $360.$
Note: Permutations and Combinations is a tricky chapter and a lot of practice is advised to be put in so as to be able to solve a question within a period of time. There are some standard formulas which need to be memorized otherwise we need to understand the logic behind every question so as to be able to proceed further and solve a variety of problems. Since factorials are involved, a lot of calculations will be encountered. So care must be taken while calculating because one mistake will lead to the wrong answer.
Complete step by step solution:
We have to divide the total number of ways in which we can write the word by the factorial of the number of times a letter is repeated.
For example we have a few letters together . Let is aabc. The combinations that can be formed are the following :
$1)$ abc
$2)$ acb
$3)$ bac
$4)$ bca
$5)$ cab
$6)$ cba
We can’t really tell whether the first a or the second a is being in these combinations. Just because we don’t know which a is in these combinations, we do not write a combination with first a and the same combinations with the second a since both of them will be the same.
In the same, repeated letter will make up the same word no matter which repetition makes up the word.
After counting we can safely say that the word is made up, $6$letters.
Let us find the number of repeated letters and the number of repetition of each letter.
We can infer from the word that the letter T is repeated for $2$times.
The non-repeating letters are P,R,E,Y.
The total number of ways to write the word “PRETTY” would be $6!$ .
Let us divide it by $2!$ so as to subtract or delete all the same words which are made up of the repetition of the repeated letter.
Upon doing so, we get the following :
$\Rightarrow \dfrac{6!}{2!}=360$
$\therefore $ The number of unique ways there are to arrange the letters in the word “PRETTY” are $360.$
Note: Permutations and Combinations is a tricky chapter and a lot of practice is advised to be put in so as to be able to solve a question within a period of time. There are some standard formulas which need to be memorized otherwise we need to understand the logic behind every question so as to be able to proceed further and solve a variety of problems. Since factorials are involved, a lot of calculations will be encountered. So care must be taken while calculating because one mistake will lead to the wrong answer.
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