
Find total number of 5 letter word that can be formed using letter of the word DRAGON
(i) Repetition allow
(ii) Repetition not allow
Answer
590.7k+ views
Hint: When repletion is allowed all the space can be filled with all the given number of objects but when repletion is not allowed every succeeding space gets 1 less object than the preceding space.
Complete step by step answer:
When repetition is allowed
Let the space for 5 letter word be
\[\begin{matrix}
\times & \times & \times & \times & \times \\
\end{matrix}\]
the $1^{st}$ space of the 5 letter word can be filled by any letter of the word DRAGON that is D, R, A, G, O and N.
Therefore,
1st space can be filled in 6 ways
Thus,
\[\begin{matrix}
6 & \times & \times & \times & \times \\
\end{matrix}\]
Now, the $2^{nd}$ space of the 5 letter word can be filled by any letter of the word DRAGON that is D, R, A, G, O and N because repetition is allowed.
Therefore,
2nd space can be filled in 6 ways
Thus,
\[\begin{matrix}
6 & 6 & \times & \times & \times \\
\end{matrix}\]
Similarly, all the other 3 space can be filled by 6 ways,
Thus,
\[\begin{matrix}
6 & 6 & 6 & 6 & 6 \\
\end{matrix}\]
Therefore,
The total number of 5 letter words with repetition will be
\[6\times 6\times 6\times 6\times 6=7776\]
When repetition is not allowed
Let the space for 5 letter word be
\[\begin{matrix}
\times & \times & \times & \times & \times \\
\end{matrix}\]
the $1^{st}$ space of the 5 letter word can be filled by any letter of the word DRAGON that is D, R, A, G, O and N.
Therefore,
1st space can be filled in 6 ways
Thus,
\[\begin{matrix}
6 & \times & \times & \times & \times \\
\end{matrix}\]
Now, the $2^{nd}$ space of the 5 letter word can be filled by only five letters of the word DRAGON because one letter is already in the word and repetition is not allowed.
Therefore,
2nd space can be filled in 5 ways
Thus,
\[\begin{matrix}
6 & 5 & \times & \times & \times \\
\end{matrix}\]
Similarly, 3rd space can be filled in 4 ways
\[\begin{matrix}
6 & 5 & 4 & \times & \times \\
\end{matrix}\]
4th space can be filled in 3 ways
\[\begin{matrix}
6 & 5 & 4 & 3 & \times \\
\end{matrix}\]
5th space can be filled in 2 ways
\[\begin{matrix}
6 & 5 & 4 & 3 & 2 \\
\end{matrix}\]
Therefore,
Total number of words that can be formed without repetition is
\[6\times 5\times 4\times 3\times 2=720\]
Note: When repetition is not allowed the number of words will be less than the words when the repetition is allowed.
Complete step by step answer:
When repetition is allowed
Let the space for 5 letter word be
\[\begin{matrix}
\times & \times & \times & \times & \times \\
\end{matrix}\]
the $1^{st}$ space of the 5 letter word can be filled by any letter of the word DRAGON that is D, R, A, G, O and N.
Therefore,
1st space can be filled in 6 ways
Thus,
\[\begin{matrix}
6 & \times & \times & \times & \times \\
\end{matrix}\]
Now, the $2^{nd}$ space of the 5 letter word can be filled by any letter of the word DRAGON that is D, R, A, G, O and N because repetition is allowed.
Therefore,
2nd space can be filled in 6 ways
Thus,
\[\begin{matrix}
6 & 6 & \times & \times & \times \\
\end{matrix}\]
Similarly, all the other 3 space can be filled by 6 ways,
Thus,
\[\begin{matrix}
6 & 6 & 6 & 6 & 6 \\
\end{matrix}\]
Therefore,
The total number of 5 letter words with repetition will be
\[6\times 6\times 6\times 6\times 6=7776\]
When repetition is not allowed
Let the space for 5 letter word be
\[\begin{matrix}
\times & \times & \times & \times & \times \\
\end{matrix}\]
the $1^{st}$ space of the 5 letter word can be filled by any letter of the word DRAGON that is D, R, A, G, O and N.
Therefore,
1st space can be filled in 6 ways
Thus,
\[\begin{matrix}
6 & \times & \times & \times & \times \\
\end{matrix}\]
Now, the $2^{nd}$ space of the 5 letter word can be filled by only five letters of the word DRAGON because one letter is already in the word and repetition is not allowed.
Therefore,
2nd space can be filled in 5 ways
Thus,
\[\begin{matrix}
6 & 5 & \times & \times & \times \\
\end{matrix}\]
Similarly, 3rd space can be filled in 4 ways
\[\begin{matrix}
6 & 5 & 4 & \times & \times \\
\end{matrix}\]
4th space can be filled in 3 ways
\[\begin{matrix}
6 & 5 & 4 & 3 & \times \\
\end{matrix}\]
5th space can be filled in 2 ways
\[\begin{matrix}
6 & 5 & 4 & 3 & 2 \\
\end{matrix}\]
Therefore,
Total number of words that can be formed without repetition is
\[6\times 5\times 4\times 3\times 2=720\]
Note: When repetition is not allowed the number of words will be less than the words when the repetition is allowed.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

