# How many words, with or without words meaning each of 3 vowels and 2 consonants can be formed from the letter of the word INVOLUTE?

Answer

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Hint: Find number of ways of selection from word “INVOLUTE”. Total number of letters is 5 multiplied by the number of arrangements.

We know the given word “INVOLUTE”.

Out of this, the total number of vowels= 4(I, O, U, E)

Total number of constants= 4(N, V, L, T)

We have to choose 3 vowels out of the 4 vowels in the word.

\[\therefore \]The number of ways to choose\[={}^{4}{{C}_{3}}-\left( 1 \right)\]

We have to choose 2 consonants out of the 4 consonants.

\[\therefore \]The number of ways to choose\[={}^{4}{{C}_{2}}-\left( 2 \right)\]

Thus, the number of ways of selecting 3 vowels and 2 consonants

\[={}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}\][from eq (1) and (2)]

We have to simplify \[{}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}\]

They are of the form \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]

\[\begin{align}

& \therefore {}^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!}=\dfrac{4!}{3!1!} \\

& {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}=\dfrac{4!}{3!2!} \\

\end{align}\]

\[\begin{align}

& {}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}=\dfrac{4!}{3!1!}\times \dfrac{4!}{3!2!} \\

& {}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}=\dfrac{4!\times 3!}{3!\times 1!}\times \dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1} \\

\end{align}\]

\[\left\{ \begin{align}

& \because 4!=4\times 3\times 2\times 1 \\

& 3!=3\times 2\times 1 \\

& 2!=2\times 1 \\

& 1!=1 \\

& 0!=1 \\

\end{align} \right\}\]

Cancel out 3! on denominator and denominator.

\[\Rightarrow \]Cancel out like terms

\[=4\times 2\times 3=24\]

\[\therefore \]The number of ways of selecting 3 vowels and 2 consonants = 24

Total number of letters = 3 vowels + 2 consonants = 5 letters

We have to arrange these 5 letters.

\[\therefore \]Number of arrangements of 5 letters \[={}^{5}{{P}_{5}}\]

This is the form \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\left\{ \begin{align}

& \because 5!=5\times 4\times 3\times 2\times 1 \\

& 0!=1 \\

\end{align} \right\}\]

\[\begin{align}

& {}^{5}{{P}_{5}}=\dfrac{5!}{\left( 5-5 \right)!}=\dfrac{5!}{0!} \\

& {}^{5}{{P}_{5}}=\dfrac{5\times 4\times 3\times 2\times 1}{1}=120 \\

\end{align}\]

Total number of words = Number of ways of selecting \[\times \]number of arrangements

Total number of words = \[24\times 120=2880\]

Total number of words \[=2880\]

Note: We use the combination \[\left( {}^{n}{{C}_{r}} \right)\] in place where the order doesn’t matter. Permutation \[\left( {}^{n}{{P}_{r}} \right)\] is used in the place where order matters.

\[\therefore \]Permutation is an ordered combination.

We know the given word “INVOLUTE”.

Out of this, the total number of vowels= 4(I, O, U, E)

Total number of constants= 4(N, V, L, T)

We have to choose 3 vowels out of the 4 vowels in the word.

\[\therefore \]The number of ways to choose\[={}^{4}{{C}_{3}}-\left( 1 \right)\]

We have to choose 2 consonants out of the 4 consonants.

\[\therefore \]The number of ways to choose\[={}^{4}{{C}_{2}}-\left( 2 \right)\]

Thus, the number of ways of selecting 3 vowels and 2 consonants

\[={}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}\][from eq (1) and (2)]

We have to simplify \[{}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}\]

They are of the form \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]

\[\begin{align}

& \therefore {}^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!}=\dfrac{4!}{3!1!} \\

& {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}=\dfrac{4!}{3!2!} \\

\end{align}\]

\[\begin{align}

& {}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}=\dfrac{4!}{3!1!}\times \dfrac{4!}{3!2!} \\

& {}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}=\dfrac{4!\times 3!}{3!\times 1!}\times \dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1} \\

\end{align}\]

\[\left\{ \begin{align}

& \because 4!=4\times 3\times 2\times 1 \\

& 3!=3\times 2\times 1 \\

& 2!=2\times 1 \\

& 1!=1 \\

& 0!=1 \\

\end{align} \right\}\]

Cancel out 3! on denominator and denominator.

\[\Rightarrow \]Cancel out like terms

\[=4\times 2\times 3=24\]

\[\therefore \]The number of ways of selecting 3 vowels and 2 consonants = 24

Total number of letters = 3 vowels + 2 consonants = 5 letters

We have to arrange these 5 letters.

\[\therefore \]Number of arrangements of 5 letters \[={}^{5}{{P}_{5}}\]

This is the form \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\left\{ \begin{align}

& \because 5!=5\times 4\times 3\times 2\times 1 \\

& 0!=1 \\

\end{align} \right\}\]

\[\begin{align}

& {}^{5}{{P}_{5}}=\dfrac{5!}{\left( 5-5 \right)!}=\dfrac{5!}{0!} \\

& {}^{5}{{P}_{5}}=\dfrac{5\times 4\times 3\times 2\times 1}{1}=120 \\

\end{align}\]

Total number of words = Number of ways of selecting \[\times \]number of arrangements

Total number of words = \[24\times 120=2880\]

Total number of words \[=2880\]

Note: We use the combination \[\left( {}^{n}{{C}_{r}} \right)\] in place where the order doesn’t matter. Permutation \[\left( {}^{n}{{P}_{r}} \right)\] is used in the place where order matters.

\[\therefore \]Permutation is an ordered combination.

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