Question

How many words with or without meaning can be formed using the letters of the word $ALLAHABAD$ so that vowels are never together?

Answer 1

First, we have to separate vowel and consonant letters of the word in ALLAHABAD.
Vowel letters are $a,e,i,o,u.$ and the consonant letters are $b,c,d,f,g,h,j,k,l,m,n,p,q,r,s,t,v,w,x,y{\text{ }}and{\text{ }}z.$
Also, we know that $ALLAHABAD$ letters contain $5$ consonant letters and $4$vowels Letters.
The consonant letters are $LLHBD$ and the vowels letters are $4$
In that condition, we are going step by step to understand the concepts.

Complete step by step answer:
The given word $ALLAHABAD$ consists of nine letters that are $4A's$ , $2L's$, $1H$, $1B$ and $1D$ .
Case a: Words formed by using all the letters
Therefore, the number of words formed by all the letters $= \dfrac{{9!}}{{4! \times 2! \times 1! \times 1!}}$
$$
= \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4! \times 2!}} \\
= 7560 \\

Case b: Now, we shall consider both L together and we need to treat LL as $1$ letter. Then, we are left with eight letters, namely $$LL$$, $$4A's$$ ,$1H$ ,$1B$ and $1D$, and we need to arrange them. Therefore, the number of words containing both L together $ = \dfrac{{8!}}{{4!}}$ $ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!}} \\\ = 1680 \\\ $ Now, we need to find the number of words that do not contain both L together. Hence, the number of words that do not contain both L together $$ = 7560 - 1680$$ $ = 5880$ Case c: The given word contains four vowels and all vowels are alike (i.e.) $$4A's$$ There exist four even places namely Also, there are 4 even places, namely $${2^{nd}}{\text{,}}{4^{th}}{\text{,}}{6^{th}}and{8^{th}}$$ . Here, we need to arrange $$4A's$$at these four even places. $ - {2^{nd}} - {4^{th}} - {6^{th}} - {8^{th}} - $ The number of ways in which $$4A's$$can be arranged at four even places $ = \dfrac{{4!}}{{4!}}$ $ = 1$ Then, we are left with five letters, namely $$2L's$$, $1H$, $1B$ and $1D$, and we need to arrange them. The number of ways in which these five letters can be arranged at the remaining five places $ = \dfrac{{5!}}{{2!}}$ $ = \dfrac{{5 \times 4 \times 3 \times 2!}}{{2!}} \\\ = 60 \\\ $ Therefore, the number of words in which vowels occupy even places $$ = (1 \times 60) = 60$$ **Note:** We have used this problem by permutation model then we already know that vowels and consonants letters $$ALLAHABAD$$ are $$5$$ consonants and $$4$$vowels and then all vowel letters are A’s. Hence, the number of words formed by all the letters $$ = 7560$$, the number of words that do not contain both L together $ = 5880$, and the number of words in which vowels occupy even places$$ = 60$$ .