# In how many arrangements of the letters of the word INTERMEDIATE 2 vowels never come together?

I went on to proceed with this problem by subtracting the number of arrangements in which all vowels were together, from the total number of arrangement which proved time consuming. An alternative solution, given by the textbook was what they called the gap method.

They find the number of ways of arranging $6$ consonants. "Then they say that there are $7$ gaps remaining (how!!!!!)" in which the remaining $6$ vowels are arranged in $\frac{7p6}{3!2!}$. The answer is $151200$

• The seven "gaps" consist of five spaces between successive consonants and two at the ends of the row $\square C \square C \square C \square C \square C \square C \square$. – N. F. Taussig May 27 '18 at 11:05

The vowels are: AEEEII.

The non-vowels are: DMNRTT.

The number of ways to arrange the vowels is $\frac{(1+3+2)!}{1!\times3!\times2!}=60$.

The number of ways to arrange the non-vowels is $\frac{(1+1+1+1+2)!}{1!\times1!\times1!\times1!\times2!}=360$.

The number of ways to choose slots for the $6$ vowels between the $6$ non-vowels is $\binom{6+1}{6}=7$.

So the number of ways to arrange this word under the given restriction is $60\cdot360\cdot7=151200$.

• $7\choose 6$=$6$?? – Upstart Jan 19 '17 at 12:02
• @Upstart: Yeah, I just fixed that... – barak manos Jan 19 '17 at 12:02

Take one easy example -

Suppose you have word UGLIER . In how many ways arranged so that vowels never comes together.

We have 5 letters and we first fill consonants (C) alternatively in them.

_ , C , _ , C , _ , C

Now we have to fill vowels. But you can notice that we can create one place at last and delete starting place. So we have 4 places to fill 3 vowels.

• So is that how that 7 came . By removing the first empty space and putting it at the end? – Sidd Jan 19 '17 at 12:54
• Yes exactly like if we put vowel on first place we have word ending with consonant. But it is not necessary that word starts with vowel. So we remove first place to start word with consonant and edit one place at last. – Kanwaljit Singh Jan 19 '17 at 13:29

how may gaps are there between the consonants..$6$ consonants so for vowels there are $7$ places.first arrange the consonants in $(6!/2)$ ways.now choose any 6 places from the 7 and arrange the 6 vowels that can be done in ($7\choose 6$$×6!)/(2!×3!)$ and finally multiply both the arrangements

• How are there 7 places . If there are a total of 12 and 6 are occupied then only 6 are supposed to be vacant right? – Sidd Jan 19 '17 at 12:46
• suppose you have 7 consonants and you have to stand besides or in between them either you stand in between places or in the extreme corners – Upstart Jan 19 '17 at 12:49
• Sorry I didn't get it yet :( Could you explain it in another way – Sidd Jan 19 '17 at 12:52
• you have two friends A and B and you have sit .Either you can sit to the left of A or in the middle of A and B or to the right of B.So total 3 choices to sit.. – Upstart Jan 19 '17 at 12:56
• But in the question above isn't there supposed to be only 6 places in which 6 letters can be filled. Like what if there are 6 chairs and 6 people then number of ways to seat them would evidently be 6p6 right. Is this analogy wrong? – Sidd Jan 19 '17 at 13:00