How many ways $INSTRUCTOR$ can be arranged such that it has two consecutive vowels?(Three consecutive vowels are not counted i.e $V1V2V3$ is not counted). Show
I was trying to solve this problem in a different way. But I got a mismatch in the answer. Following is my solution INSTRUCTOR has three vowels $I,U$ and $O$. Lets arrange the three vowels like this _ V1V2 _ V3 _ The above representation shows V1,V2,V3 as vowels and "_" as consonants. We have total 7 consonants which can fill all these three holes(refereeing to "_"). There must be at least one consonant between V1V2 and V3, else three vowels will come together. It can be written like this: $X_1+X_2+X_3=7$ $X_1\ge0,X_2\ge1,X_3\ge0$ This will give result $C(8,2)=28$ We can arrange these 3 vowels in $3!$ ways. Again we can also interchange the position of $V_1,V_2$ and $V_3$ like this. _V3_V1V2_ So we have total $(28)(3!)(2)$ ways of arranging three vowels. Now the other 7 consonants can be arranged in $\frac{(7!)}{(2!)(2!)}$ ways. So answer should be $$\frac{(28)(3!)(2)(7!)}{(2!)(2!)}$$ Given answer $$\frac{(64)(3!)(7!)}{(2!)(2!)}$$ What's wrong here? Help appreciated :)
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Fraction consists of 8 letters which has 3 vowels (a, I, o) and rest 5 are consonants. It can be -F- R- C- T- N so that no 2 vowels are together. Now 3 vowels can fill these 6 places in (6,3) ways.
What fraction of letters are vowels?Two of those letters, or 2/6 of the letters, are vowels, and four of the letters, or 4/6, are consonants.
How many different words can be formed from the letter combine so that no two vowels are together?The three vowels can be arranged in three ways: AAE, AEA, EAA. Hence, the number of such arrangements is 5!
How many arrangements are there where no two vowels are next to each other?ways. In total we have (63)×3! ×5! =14400 ways.
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