Question

Find the number of 5 letter words that can be formed using the letter of the word NATURE that begins with N when repetition is allowed.

Answer 1

Hint: Here we count the total number of letters in the word NATURE. Then we find possibilities of 5 letter words that can be formed by fixing the first letter as N and finding the number of letters that can be placed at second place, third place, fourth place and fifth place respectively. We use the concept that when repetition is allowed, we can choose from a total number of letters for every place.

Complete step-by-step answer:
We are given the word NATURE.
Number of letters in the word NATURE are 6.
We have to form five letter words that start with the letter N. So we fix N at the place of the first letter and find ways to fill the other four positions by 6 different letters simultaneously.
Since, repetition is allowed, so each place has 6 possibilities of choosing a letter.
Number of ways to fill the first place is 1. (Because letter can only be N)
Number of ways to fill the second place is 6. (Because letter can be any of the six letters)
Number of ways to fill the third place is 6. (Because letter can be any of the six letters)
Number of ways to fill the fourth place is 6. (Because letter can be any of the six letters)
Number of ways to fill the fifth place is 6. (Because letter can be any of the six letters)
Number of 5 letter words that can be formed using letters of NATURE where the word starts from N is given by multiplication of the number of ways to fill each place.
\[ \Rightarrow 1 \times 6 \times 6 \times 6 \times 6\]
\[ \Rightarrow 1296\]
Therefore, the number of 5 letter words that can be formed using letters of NATURE where the word starts from N is 1296.

Note: Alternate method:
Since we have to form a five letter word where the first letter is fixed as N, we use combination formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\] where n is the total number of letters we are choosing from and r is the number of letters we have to choose.
We find possibilities of filling each place simultaneously.
For first position we have n as 1 and r as 1 because we have only one letter N.
Number of ways to fill first position is \[^1{C_1} = \dfrac{{1!}}{{(1 - 1)!1!}}\]
\[{ \Rightarrow ^1}{C_1} = \dfrac{{1!}}{{0!1!}}\]
\[{ \Rightarrow ^1}{C_1} = 1\]
No we find number of possibilities for remaining four positions.
Since number of letters we have from the word NATURE is 6
\[ \Rightarrow n = 6\]
Also, the number of letters that we have to choose for each place is 1
\[ \Rightarrow r = 1\]
So, number of ways to fill each position is
\[{ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{(6 - 1)!1!}}\]
\[{ \Rightarrow ^6}{C_1} = \dfrac{{6!}}{{5!1!}}\]
Now we know factorial opens up as \[n! = n(n - 1)!\]
\[{ \Rightarrow ^6}{C_1} = \dfrac{{6 \times 5!}}{{5!1!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^6}{C_1} = 6\]
Since we have 4 positions to be filled, therefore number of letters that can be formed will be given by \[^1{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}\]
Substituting the values we get
\[{ \Rightarrow ^1}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1} = 1 \times 6 \times 6 \times 6 \times 6\]
\[{ \Rightarrow ^1}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1}{ \times ^6}{C_1} = 1296\]