Solution : No. of letters in the word 'DAUGHTER' =8
[i] If there is no restriction,
No. of permutations to fill 8 places with given 8 letters `= P_[8] = 8! = 40320`.
[ii] In the word 'DAUGHTER'. Vowels = A,U,E and Consonants = D,G,T,H,R
`rArr` No. of vowels and consonants are 3 and 5 respectively.
`:'` A,U,E are always together
`:.` Treating these letters as one letter, the letters are `5 +1 =
6`
Now the number of arrangement of 6 letters at 6 places `= .^[6]P_[6] = 6! = 720` and Number of arrangement of letters A,U,E at 3 places `= .^[3]P_[3] = 3! =6`.
`:.` Required arrangements `= 720 xx 6 = 4320`
[iii] For the words starting with A
A will be at first place and no of arrangements of remaining 7 letters at remaining 7 places
`=.^[7]P_[7] = 7! = 5040`.
[iv] For the words starting with A and ending with R,
No.
of arrangements of remaining 6 letters at remaining 6 places
`=.^[6]P_[6] = 6! = 720`.
Hi all,
I would appreciate any feedback on my solution to this problem.
Problem:
How many arrangements of the word DAUGHTER are there if none of the vowels can ever be together?
Solution attempt:
3 vowels and 5 consonants
Arrangements with no restrictions = 8! = 40320All [three] vowels together = 6!3!=4320Two vowels together = 7!2!=10080
So,
No vowels together = No restrictions - three vowels together - two vowels together=40320 - 4320 - 10080 = 25920 ways
Does my logic make sense? Have I missed any cases? I'm trying to use permutations only.
Thanks for the help!
Edit:
Thanks for all the help!
The gist I was getting from the comments was that when I subtract 2 vowels together I am also getting 'more than 2 vowels together' [so, three vowels]. To compensate for this I should add the ways of arranging 3 vowels together back so as to not subtract twice.
No restrictions - 2 vowels together + 3 vowels together = 8! - 7! * 6 + 6!3! = 14400
How many words can be formed from the letters of the word ‘DAUGHTER’ so that
[i] The vowels always come together?
[ii] The vowels never come together?
Answer
Verified
Hint: The word daughter has $8$ letters in which $3$ are vowels. For the vowels to always come together consider all the $3$ vowels to be one letter [suppose V] then total letters become $6$ which can be arranged in $6!$ ways and the vowels themselves in $3!$ ways.Complete step-by-step answer:
Given word ‘DAUGHTER’ has $8$ letters in which $3$ are vowels and 5 are consonants. A, U, E are vowels and D, G, H, T, R are consonants.
[i]We have to find the total number
of words formed when the vowels always come together.
Consider the three vowels A, U, E to be one letter V then total letters are D, G, H, T, R and V. So the number of letters becomes $6$
So we can arrange these $6$ letters in $6!$ ways. Since the letter V consists of three vowels, the vowels themselves can interchange with themselves. So the number of ways the $3$vowels can be arranged is $3!$
Then,
$ \Rightarrow $ The total number of words formed will be=number of ways the $6$
letters can be arranged ×number of ways the $3$ vowels can be arranged
On putting the given values we get,
$ \Rightarrow $ The total number of words formed=$6! \times 3!$
We know $n! = n \times \left[ {n - 1} \right]! \times ...3,2,1$
$ \Rightarrow $ The total number of words formed=$6 \times 4 \times 5 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1$
On multiplying all the numbers we get,
$ \Rightarrow $ The total number of words formed=$24 \times 5 \times 6
\times 6$
$ \Rightarrow $ The total number of words formed=$120 \times 36$
$ \Rightarrow $ The total number of words formed=$4320$
The number of words formed from ‘DAUGHTER’ such that all vowels are together is $4320$.
[ii]We have to find the number of words formed when no vowels are together.
Consider the following arrangement- _D_H_G_T_R
The spaces before the consonants are for the vowels so that no vowels come together. Since there are $5$ consonants so they can be
arranged in $5!$ ways.
There are $6$ spaces given for $3$ vowels. We know to select r things out of n things we write use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$
So to select $3$ spaces of out $6$ spaces =${}^6{{\text{C}}_3}$
And the three vowels can be arranged in these three spaces in $3!$ ways.
$ \Rightarrow $ The total number of words formed=${}^6{{\text{C}}_3} \times 3! \times 5!$
$ \Rightarrow $ The total number of words
formed=$\dfrac{{6!}}{{3!6 - 3!}} \times 5! \times 3!$
$ \Rightarrow $ The total number of words formed=$\dfrac{{6!}}{{3!}} \times 5!$
On simplifying we get-
$ \Rightarrow $ The total number of words formed=$\dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} \times 5!$
$ \Rightarrow $ The total number of words formed=$120 \times 5 \times 4 \times 3 \times 2 \times 1$
On multiplying we get,
$ \Rightarrow $ The total number of words formed=$14400$
The total number of words
formed from ‘DAUGHTER’ such that no vowels are together is $14400$.
Note: Combination is used when things are to be arranged but not necessarily in order. Permutation is a little different. In permutation, order is important. Permutation is given by-
$ \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ Where n=total number of things and r=no. of things to be selected.