Question

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 1

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

\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.