Question

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 1

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.