Question

Find the number of words that can be formed by using all letters of the word ' DAUGHTER'. If:
(i) Vowels occur in first and last place.
(ii) Start with the letter G and end with the letters H.
(iii) Letters G, H, T always occur together.
(iv) No two letters of G, H, T are consecutive
(v) Not all vowels occur together
(vi) Vowels always occupy even places.
(vii) Order of vowels remains the same.
(viii) Relative order of vowels and consonants remains the same.
(ix) Number of words is possible by selecting 2 vowels and 3 consonants.

Answer 1

Hint: In the above question, we are asked to write the numbers of words that can be formed using all the letters of the word “DAUGHTER” and in each of the options the conditions for framing the words are different. We will start by counting the total letters in the word ‘Daughter’ and then we will sort out the vowels and consonants. After this, we begin solving the subparts of the question step by step.

Complete step-by-step solution:
GIven: 'DAUGHTER': Total letters are 8.
Vowels are 'AUE'
Consonants are 'DGHTR'
(i) Vowels occur in first and last place.
As it is mentioned that vowels occur in the first and last place of the words so for the first letter, there are 3 possibilities. After filling the first place of the word, we have two possibilities left so in the last place of the word 2 possibilities are there.
Now, from the 8 places, 2 places (first and last) are occupied then we are left with 5 blank places so the filling of these 6 places are done in 6! ways because all the remaining 6 letters are different and we know that arranging 6 different items in a row is given by $6!$.
So, the number of ways is equal to:
 \[\begin{align}
  & 3\times 2\times 6! \\
 & =3\times 2\times 6\times 5\times 4\times 3\times 2\times 1=4320 \\
\end{align}\]
(ii) Start with the letter G and end with the letter H.
In this part, we have fixated the first and last place with G and H respectively so the remaining places to be filled are 6 and that can be filled by arranging the 6 different letters in 6 blanks which is equal to:
  \[\begin{align}
  & 6!=6\times 5\times 4\times 3\times 2\times 1 \\
 & =720 \\
\end{align}\]
Hence, the total number of ways is equal to 720.
(iii) Letters G, H, T occurs together always
It is given that G, H, T occurs together always so we are considering here GHT as one object.
From the 8 letters, now 3 letters always occur together so we are left with 5 letters so the total number of objects becomes equal to:
 \[5+1=6\]
So, the number of ways of arranging them is $6!$.
Despite it is given that GHT always occurs together but it is not given that it always occurs in the form GHT where G occurs first followed by H and followed by T so there is an internal arrangement of GHT will occur and the number of possible internal arrangements is $3!$ ways.
So, the total number of ways is equal to:
 \[\begin{align}
  & 3!\times 6!=3\times 2\times 1\times 6\times 5\times 4\times 3\times 2\times 1 \\
 & =4320 \\
\end{align}\]
(iv) No two letters of G, H, T are consecutive.
To find the above number of possibilities, we are going to subtract the cases when two of the letters G, H, T are consecutive from the total number of words that can be formed from the word “DAUGHTER”.
The total number of words that can be from the word “DAUGHTER” is $8!$ because all the 8 letters are different then the possible arrangements of 8 different letters is $8!$.
Now, to find the possibilities when no two letters are consecutive can be found by taking any two of the three letters G, H, T, will come together for that we have to first select any two letters from the three letters G, H, T which is equal to ${}^{3}{{C}_{2}}$ and when any two of the three letters G, H, T, will come together then the number of objects become 7 and arrangements of those objects is $7!$. Now, the internal arrangement of the two letters is also there which is equal to $2!$. The total number of ways when any two of G, H, T is consecutive:
$\begin{align}
  & {}^{3}{{C}_{2}}7!2! \\
 & =\dfrac{3!}{2!1!}7!2! \\
 & =3.2.1.7.6.5.4.3.2.1 \\
 & =30240 \\
\end{align}$
The total permutations of 8 different letters are equal to:
 $\begin{align}
  & 8! \\
 & =8.7.6.5.4.3.2.1 \\
 & =40320 \\
\end{align}$
Subtracting 30240 from 40320 we get,
$\begin{align}
  & 40320-30240 \\
 & =10,080 \\
\end{align}$
Hence, there 10,080 words are possible.
(v) Not all vowels occur together.
The above number of words can be found by writing the number of possibilities when all the vowels come together and then subtract the result of these possibilities from the total permutations of 8 letters.
There are three vowels present i.e. A, U, E. So, when they occur together then AUE will be treated as one object and then the number of objects becomes 6. This is similar to the option (iii) so the total number of words that could be possible when the three vowels are together is equal to:
$6!3!$
Total permutations are equal to 8!. Subtracting permutations when the three vowels are together from the total permutations we get,
$\begin{align}
  & 8!-6!3! \\
 & =8.7.6.5.4.3.2.1-6.5.4.3.2.1.3.2.1 \\
 & =40320-4320 \\
 & =36000 \\
\end{align}$
Hence, the total number of words that can be formed from the word “DAUGHTER” such that not all vowels are together is 36000.
(vi) Vowels occupy even space.
There are 4 even places from 8 places and we have 3 vowels.
'So we can arrange vowels in \[{}^{4}{{P}_{3}}\] ways. The remaining 5 letters can be arranged in 5! ways.
So, the number of ways is equal to:
 \[\begin{align}
  & 5\times {}^{4}{{P}_{3}}=\dfrac{4!}{1!}\times 5! \\
 & =4\times 3\times 2\times 1\times 5\times 4\times 3\times 2\times 1=2880 \\
\end{align}\].
(vii) Order of vowels remains the same.
In this part, we have to find the possible number of words such that, places of vowels remain the same meaning the places occupied by the three vowels in the word “DAUGHTER” remain the same. So, we cannot touch the three places in the 8 places so we are left with 5 places and arranging 5 letters in 5 places is 5! And we cannot arrange the vowels because the order of the vowels remains the same so the answer is $5!$ and expanding this factorial will give us:
 $\begin{align}
  & 5! \\
 & =5.4.3.2.1 \\
 & =120 \\
\end{align}$
Hence, the total number of ways is equal to 120.
(viii) The order of vowels and consonants remains the same.
 We have to find the number of possible words formed from the letters of the word “DAUGHTER” such that the order of vowels and consonants remain the same. As the word “DAUGHTER” contains vowels and consonants only and its order remains the same means there is only one word possible which is the word itself “DAUGHTER” so only 1 word is possible.
(ix) We need to select two vowels which can be done in \[{}^{3}{{C}_{2}}\] ways. Selecting three consonants can be done in \[{}^{5}{{C}_{3}}\] ways. Now, the word we require contains only five letters so arranging 5 different letters is equal to $5!$.
So, the number of words that are possible by selecting 2 vowels and 3 consonants is equal to:
\[\begin{align}
  & {}^{3}{{C}_{2}}\times {}^{5}{{C}_{3}}\times 5! \\
 & =3\times 10\times 5\times 4\times 3\times 2\times 1=3600 \\
\end{align}\]
Hence, 3600 words can be formed by selecting 2 vowels and 3 consonants from the 8 letters of the word “DAUGHTER”.

Note: We have to remember the formula of combination \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\times (n-r)!}\] and formula of permutation \[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\] to find the number of ways which is a critical part to solve the question. We have to read this type of question 4 to 5 times to clearly understand it. We can make a mistake in applying the permutation and combination formulas.