Question

Find the number of 4 letter words, with or without meaning which can be formed out of letters of the word, NOSE
(i) Repetition of the letter is not allowed
(ii) Repetition is allowed

Answer 1

Hint: We will solve this question by using a filling up space method. In this method we will fill the spaces which will be given as $-,-,-,-$ one by one starting from left to right by the number of choices that we will get for each place. This method is much easier than others. Therefore with the help of this method we will find the required possibilities.

Complete step-by-step answer:
According to the question we are given the word NOSE which consists of 4 letters. Out of these 4 letters we are supposed to form 4 letters word only but in a jumbled form means it can be with or without meaning.
Now, we will first consider the case (i) in which we are given a limitation which says that the repetition of the letters is not allowed. For this we will consider places like this $-,-,-,-$. This will make solving easy.
Now as we have 4 letters and the first place can be filled by 4 letters. Therefore, first place has 4 choices. This is done as $4,-,-,-$.
Now we consider the second place. As the repetition is not allowed therefore we are now left with 3 choices. Therefore, the second place is filled by 3 choices. That is $4,3,-,-$.
Similarly, for the third we have 2 choices and for the last one we have only one choice. Thus we now have $4,3,2,1$.
Now we only need to multiply them by each other and we will have total possibilities for this case. Thus, we have $4\times 3\times 2\times 1=24$. Hence, the total number of possibilities is 24 here.
Now we will consider the case (ii) in which we are given a limitation which says that the repetition of the letters is allowed. For this we will consider $-,-,-,-$.
Now as we have 4 letters and the first place can be filled by 4 letters. Therefore, first place has 4 choices. This is done as $4,-,-,-$.
Now we consider the second place. As the repetition is allowed therefore we still have 4 choices. Therefore, the second place is filled by 4 choices. That is $4,4,-,-$.
Similarly, for the third we have 4 choices and for the last one we have 4 choices again. Thus, we now have $4,4,4,4$.
Now, we only need to multiply them by each other and we will have total possibilities for this case. Thus, we have $4\times 4\times 4\times 4=256$. Hence, the total number of ways is 256 here.

Note: Alternate method of solving this question when the repetition is not allowed is given below.
We can use the concept of permutation here. The formula of permutation is given by $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ where n are the total number of objects and r is the selected objects. Therefore in case (i) we have repetition not allowed. Thus the number of possibilities it will have is given by
$\begin{align}
  & P_{4}^{4}=\dfrac{4!}{\left( 4-4 \right)!} \\
 & \Rightarrow P_{4}^{4}=\dfrac{4!}{\left( 0 \right)!} \\
\end{align}$
As $0!=1$ therefore, we have
$\begin{align}
  & P_{4}^{4}=\dfrac{4!}{1} \\
 & \Rightarrow P_{4}^{4}=4\times 3\times 2\times 1 \\
 & \Rightarrow P_{4}^{4}=24 \\
\end{align}$
Hence, total possibilities in case (i) are 24.