Question

# How many different words can be formed from the letters of the word GANESHPURI when:How many words of 5 letters each can be formed each containing $3$ consonants and $2$ vowels?

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Hint: First, figure out how many consonants and vowels are there in the word GANESHPURI and then write the consonants and the vowels separately. Use the concept of permutation and combination to select $3$ letters from consonants and $2$ letters from vowels.

Before proceeding with the question, we must know the important formulas of permutation and combination that will be used to solve this question.
Let us assume we have $n$ different letters from which we have to select $r$ letters where $r${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.....................\left( 1 \right)$If we have a word which contains$n$different letters in it, then the number of ways in which we can rearrange these letters is equal to$n!...........\left( 2 \right)$. In this question, we are given the word GANESHPURI. From this word, we have to find the number of words that are having$3$consonants and$2$vowels. Let us first separate out the consonants and the vowels from the word GANESHPURI. In the word GANESHPURI, the consonants are G, N, S, H, P, R. So, there$6$consonants in the word GANESHPURI. In the question, it is given that we have to select$3$consonants out of these$6$consonants. Substituting$n=6$and$r=3$in the formula$\left( 1 \right)$, the number of ways of selecting$3$consonants is,$\begin{align}
& {}^{6}{{C}_{3}}=\dfrac{6!}{3!\left( 6-3 \right)!} \\
& \Rightarrow \dfrac{6!}{3!3!} \\
& \Rightarrow \dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 3\times 2\times 1} \\
& \Rightarrow 20............\left( 3 \right) \\
\end{align}$In the word GANESHPURI, the vowels are A, E, U, I. So, there are$4$vowels in the word GANESHPURI. In the question, it is given that we have to select$2$vowels out of these$4$vowels. Substituting$n=4$and$r=2$in the formula$\left( 1 \right)$, the number of ways of selecting$2$consonants is,$\begin{align}
& {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!} \\
& \Rightarrow \dfrac{4!}{2!2!} \\
& \Rightarrow \dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1} \\
& \Rightarrow 6............\left( 4 \right) \\
\end{align}$The number of words that are having$3$consonants and$2$vowels can be found out by multiplying equation$\left( 3 \right)$and equation$\left( 4 \right)$and are equal to $20\times 6=120$. So, the number of ways of selecting these$5$letter words is equal to$120$. Also, all these words can be rearranged within themselves. The number of words that can be formed by rearranging a single$5$letter word can be found out by substituting$n=5$in equation$\left( 2 \right)$and are equal to$5!=5\times 4\times 3\times 2\times 1=120$. As for each of the$120$selected words, there are$120$rearrangements possible, therefore, the total number of words is equal to$120\times 120=14400$. Note: There is a possibility of committing a mistake while calculating the final answer. There is a possibility that one does not rearrange the$5$letter words which we got after selecting$3$consonants and$2$vowels. So, there is a possibility that one does not multiply$120$by$5!\$ which will lead us to an incorrect answer.