Arrangement of the word DIAGONISE such that each word start with a vowel and ends with a consonant

Question

How many arrangements can be made using all the letters of the word DIAGONISE such that each of these starts with a vowel and ends with a consonant?

Answer 1

Apologies, my previous answer was incorrect...

Basically, since I appears twice -- you need to separate it into two case statements:

CASE 1: (I is picked as the first vowel)

This will leave you with 4 ways to pick the last consonant, and then $7!$ combinations, (since the elements that we are permutating are now unique! So there is no need to divide by 2 here.

CASE 2: (I is not picked as the first vowel)

In other words, there are only 3 vowels to choose from (since we have already handled the others in Case 1 ). There are still 4 ways to pick the last consonant, and then $\frac{7!}{2}$ combinations (since you now have two identical elements in your 'body of the word')

Add up both of these cases:

Answer 2

No of vowels = 2I + A + O + E= 5ways

No of Consonants = D + G + N + E =4ways

Remaining words can be arrange in $7!$

so,,result will be

$$\frac{(5)(4)(7!)}{(2!)}$$