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 ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

& \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{aligned}$$ $$\begin{aligned} & {}^{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{aligned}$$ $$\left\\{ \begin{aligned} & \because 4!=4\times 3\times 2\times 1 \\\ & 3!=3\times 2\times 1 \\\ & 2!=2\times 1 \\\ & 1!=1 \\\ & 0!=1 \\\ \end{aligned} \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{aligned} & \because 5!=5\times 4\times 3\times 2\times 1 \\\ & 0!=1 \\\ \end{aligned} \right\\}$$ $$\begin{aligned} & {}^{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{aligned}$$ Total number of words = Number of ways of selecting $$\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.