Question

In how many ways can the letters of the word DIRECTOR be arranged so that the three vowels are never together?

Answer 1

Hint: Find the total number of words which can be formed using the letters of the word director. This is equivalent to the number of permutations of 8 objects in which two are alike of one kind and the rest all are different. Use the fact that the number of permutations of n different things in which p are alike of one kind and the rest r are different is given by $\dfrac{n!}{p!}$.

Complete step-by-step solution -
Now find the number of the ways in which the Vowels occur together. This can be done by considering the group of vowels as a single letter and then finding the number of permutations in which these letters can be arranged. Finally, take into account the internal arrangement of the group of the vowels and hence find the number of words in which all three vowels occur together. The number of words in which all three vowels do not occur together is equal to the number of words in which all three vowels occur together subtracted from the total number of words.
We have the letters of the word DIRECTOR are
{'D': 1, 'I': 1, 'R': 2, 'E': 1, 'C': 1, 'T': 1, 'O': 1}
Consider the two Rs as different$\left( {{R}_{1}},{{R}_{2}} \right)$.
The total number of words which can be formed from these letters is then equal to $8!$. But since actually, the Rs are identical, if we interchange the position of ${{R}_{1}}$ and ${{R}_{2}}$, no new word is formed.
Hence in the above calculation, each word is calculated twice.
Hence the total number of words which can be formed using the letters of the word DIRECTOR $=\dfrac{8!}{2!}$
Now consider as if all the vowels are single letter( Say letter IEO)
Hence, we have the letter D(1), R(2), C(1), T(1), IEO(1)
The number of words which can be formed from these letters following a similar argument as above $=\dfrac{6!}{2!}$.
But the letters in the letter IEO can internally arrange in 3! ways. Hence the total number of words in which vowels are together $=\dfrac{6!}{2!}\times 3!=3\times 6!$
Hence, the total number of words in which all three vowels do not occur together $=\dfrac{8!}{2!}-3\times 6!=18000$.

Note: Alternative Solution:
Finding the number of words in which no vowels are together:
We arrange all the letters except the vowels. This can be done in $\dfrac{5!}{2!}$ ways.
Now in between these 5 letters, we have 6 gaps. Three gaps can be selected in $^{6}{{C}_{3}}=20$ ways, and in these three gaps vowels can be filled in $3!=6$ ways.
Hence the total number of e in which no vowels words together $60\times 20\times 6=7200$
Find the number of words in which I and E are together
We arrange all the letters except the vowels. This can be done in $\dfrac{5!}{2!}$ ways.
Now in between these 5 letters, we have 6 gaps. Two gaps can be selected in $^{6}{{C}_{2}}=15$ ways, and in these two gaps vowels(O and IE) can be filled in $2!=2$ ways. The letters of the letter IE can internally arrange in $2!=2$ ways
Hence the total number of words in which vowels I and E are together $60\times 15\times 2\times 2=3600$
Similarly, the total number of words in which I and O are together = 3600 and the number of words in which E and O are together = 3600
Hence, the total number of words in which all three vowels do not occur together $=3600+3600+3600+7200=18000$, which is the same as obtained above.