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# 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? Verified
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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.