Question

Find the different permutation of the letters of the word BANANA.

Answer 1

Hint: To solve the given question, we need to know how to find the different permutations of n objects when there are ${{n}_{1}}$ repeated items, ${{n}_{2}}$ repeated items, ……${{n}_{k}}$ repeated items. This is represented by the general formula, $P=\dfrac{n!}{{{n}_{1}}!{{n}_{2}}\ldots {{n}_{k}}!}$, where n stands for the total items and ${{n}_{1}},{{n}_{2}},{{n}_{k}}$ stand for the items that are being repeated. We have the word given to us as BANANA and we know that the word contains repeating words, which are A and N, so we will find the permutation using the formula mentioned above.

Complete step-by-step answer:
We have been asked to find the different permutations of the letters of the word BANANA. We can see from the word that there are some letters that are repeating in it. The letters that are repeating are letter A and letter N. Now, we know that we can find the permutation of objects when there are repeating items in it by using the general formula, $P=\dfrac{n!}{{{n}_{1}}!{{n}_{2}}\ldots {{n}_{k}}!}$ where is the total items and ${{n}_{1}},{{n}_{2}},{{n}_{k}}$ are the items that are being repeated. So, let us consider the word given to us: BANANA. We can see there that the total number of letters in the word is 6 and the number of the repeating letters A is 3 and B is also 3. So, applying it in the formula, we have $n=6,r=3,q=2$. So, we get,
$\begin{align}
  & {}^n{{P}_{r,q}}=\dfrac{n!}{r!q!} \\
 & \Rightarrow {}^{n}{{P}_{r,q}}=\dfrac{6!}{2!3!} \\
 & \Rightarrow {}^{n}{{P}_{r,q}}=\dfrac{1\times 2\times 3\times 4\times 5\times 6}{\left( 1\times 2 \right)\left( 1\times 2\times 3 \right)} \\
 & \Rightarrow {}^{n}{{P}_{r,q}}=60 \\
\end{align}$
Hence, we get the permutations of the letters in the word BANANA as 60.

Note: If we were given the word where all the letters were different, then we could have used the formula, $n{{P}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ , but we have used the formula, ${}^{n}{{P}_{r,q}}=\dfrac{n!}{r!q!}$, since the word had letters that were repeating. Now, if we consider any two combinations from the word BANANA, say ABN and BNA, then we can use any of the 3 A’s of the word and any of the 2 N’s of the word and it makes the same combination as above. SO, we understand here that the number of permutations are lesser than the actual for a word that has repeating letters.