Combination Problem: Arranging letters of word DAUGHTER

Question

The number of ways in which we can form a 8 letter word from the letters of the word DAUGHTER such that all vowels are never occur together is

My approach:

As they are 5 consonants(DGHTR) and 3 vowels(AUE) we can arrange 5 consonants in 5! ways. $$*D*G*H*T*R*$$ Now 6 empty slots can be filled with either all vowels separately (A,U,E), which can be done in

$\binom{6}{3}\times3!$ ways

or 6 empty slots can be filled with a vowel and 2 vowel occurring together (A,UE or U,AE or E,AU) which can be done in

$\binom{6}{2}\times2!+\binom{6}{2}\times2!+\binom{6}{2}\times2!=3\times\binom{6}{2}\times2!$

Summing up all the possibilities = $5!\times(\binom{6}{3}\times3!+3\times\binom{6}{2}\times2!)=25200$

But answer is given as 36000, What possibility I missed out.

Answer 1

I don't know what possibility you missed out, but I'd think it's easier to calculate how many possibilities there are where all the vowels do appear together, and subtract that from the total $8!=40\,320$.

If all the vowels appear together, then we can think of those three together as one letter, which means there are $6!$ different words we can write. But for each of those there are $3!$ ways of rearranging the vowels within the block. So there are $6!\cdot 3! = 4\,320$ different words where all the vowels appear together.

This means there are $40\,320 - 4\,320 = 36\,000$ words where all the vowels do not appear together.

Answer 2

You haven't included the possibility of permuting e.g. the U and E in UE when they're together (and the same in the other two cases).

When this is included, your method works:

$$5! \times \left(\binom{6}{3} \times 3!+3 \times \binom{6}{2} \times 2! \times 2! \right)=36000.$$

Answer 3

You have forgotten to multiply by 2 in the last term because both the vowels that are together can be switched as well as the order of vowel placement.

I think the better way to do this (As mentioned by Arthur) is to consider the complement i.e. think of (A,E,U) as a letter. Then the letters are (D,G,H,T,R,AEU) and the total number of ways of arranging them is 6!. Also (A,E,U) can be arranged in 3! ways so the complement is 6!3!. Also the total number of ways of arranging 8 letters is 8! so your answer is 8!-6!3!=36,000.

Answer 4

ANSWER:

Since we are arranging 8 letters:

The sample space, n(s) = 8! = 40 320

If vowels are placed next to each other, then we must treat them as a single letter.

Treating vowels as single letter means we now have 6 letters to arrange.

Number of arrangements = 6! 3! Thus, No. of arrangements = 4 320

NOW: If vowels are NOT placed next to each other, the number of arrangements will be the difference between the above results.

Number of arrangements when vowels are NOT placed to each other = Sample space - No. of arrangements when vowels are placed next to each other.

No. of arrangements = 40 320 - 4 320

No. of arrangements = 36 000.