     Question Answers

# 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?  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.
View Notes
Biology Root Words Starting with Ab or Abs  Biology Root Words Starting With Hyper  Biology Root Words Starting with Cide  Biology Root Words Starting with ‘Geno’  Numbers in Words  Words  Choice of Words  Root Words For Poly  Root Words For Macro  Root Words for Aqua  