How many different letter arrangements can be made from the letter of the word extra in such a way that the vowels are always together 48 60 40?

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Since there are $5$ vowels over $11$ letters, we first arrange them so that we have no adjacent vowels.

We distinguish four classes of arrangements (here $X$ is a place for a vowel and $Y$ is a place for a consonant):

(i) $1$ with no adjacent consonants: $$YXYXYXYXYXY$$

(ii) $4$ with 3 adjacent consonants: $$X\color{blue}{YYY}XYXYXYX, XYX\color{blue}{YYY}XYXYX, XYXYX\color{blue}{YYY}XYX, XYXYXYX\color{blue}{YYY}X$$

(iii) $10$ with 2 adjacent consonants: $$\color{blue}{YY}XYXYXYXYX,YX\color{blue}{YY}XYXYXYX,YXYX\color{blue}{YY}XYXYX, YXYXYX\color{blue}{YY}XYX, YXYXYXYX\color{blue}{YY}X, X\color{blue}{YY}XYXYXYXY,XYX\color{blue}{YY}XYXYXY,XYXYX\color{blue}{YY}XYXY, XYXYXYX\color{blue}{YY}XY, XYXYXYXYX\color{blue}{YY}$$

(iv) $6$ with 2 couples of adjacent consonants: $$X\color{blue}{YY}X\color{blue}{YY}XYXYX, X\color{blue}{YY}XYX\color{blue}{YY}XYX, X\color{blue}{YY}XYXYX\color{blue}{YY}X,XYX\color{blue}{YY}X\color{blue}{YY}XYX,XYX\color{blue}{YY}XYX\color{blue}{YY}X,XYXYX\color{blue}{YY}X\color{blue}{YY}X$$

Hence, the total number of valid arrangements is $$\underbrace{\frac{5!}{2}}_{\text{vowels}}\cdot \left(\underbrace{1\cdot\left(\frac{6!}{2!3!}\right)}_{\text{case (i)}}+ \underbrace{4\cdot31}_{\text{case (ii)}}+ \underbrace{10\cdot\left(\frac{6!}{2!3!}-16\right)}_{\text{case (iii)}} +\underbrace{6\cdot\left(\frac{6!}{2!3!}-28\right)}_{\text{case (iv)}} \right)=48960.$$ Please fill the details of the above computation and let me know if you need further help.