Find the number of permutations of the word "AUROBIND" in which vowels appear in alphabetical order? I thought of 2 ways to solve this

*

*Firstly put all vowels in alphabetical order  A_I_O_U and for the remaining $5$ gaps each consonant has $5$ choices to go to. So total choices becomes $5^4 \cdot 4!$.


*Firstly put all vowels in alphabetical order  A_I_O_U and for the remaining 5 gaps we need to solve the equation $g_1+g_2+g_3+g_4+g_5=4$. And the number of solutions to this equation is the answer. That is $8C4 \cdot 4!$.
I don't know which one is correct and why other one is wrong.
 A: If you first place the vowels in order, $-A-I-O-U-$
the first consonant can be placed at any of $5$ gaps,
but each time a letter is placed, an extra gap is created for the next one
thus # of permutations $= 5\cdot6\cdot7\cdot8 = 1680$
Your mistake in the first method is that you instead took $5^4,$
and also multiplied by $4!$ permutations of the vowels
Your second method is correct

ADDED ANOTHER WAY

*

*Without any restrictions, there are $8!$ permutations.


*The vowels can be permuted in 4! ways, but only one  is in alphabetical order.


*Thus the desired number of permutations $=\dfrac{8!}{4!}=1680$
__
A: Another way to think about it:

*

*We begin with $8$ empty boxes in a row.

*We choose $4$ boxes to add the vowels to, and since the vowels must be in order, there are $\binom{8}{4}$ ways to do this.

*For each possibility, we are left with $4$ consonants which are to be added to the remaining $4$ boxes, and since they can be in any order, we can do this in $4!$ ways.

Thus we get $\binom{8}{4} 4! = \frac{8!}{4!}$ ways.
A: Your second attempt is correct.
To see what is wrong with your first attempt, consider the arrangement BADINORU.  By choosing to place the B in the first open space, the D in the second open space, the N in the third open space, and the R in the fourth open space, you have already arranged the four consonants.  Multiplying by $4!$ then arranges the consonants a second time.  It only makes sense to multiply by $4!$ if all four consonants are placed in the same space.  For all other arrangements, you are multiplying by too large a factor.
In your second attempt, you simply decided how many consonants would be placed in each gap, then arranged the consonants.  In doing so, you only arranged the four consonants once, thereby avoiding the overcounting in your first attempt.
The second method provided by true blue anil shows how to correct your first attempt.
We could also approach the problem by placing the consonants first.  There are eight positions to fill, so we have eight ways to place the B, which leaves seven ways to place the D, six ways to place the N, and five ways to place the R. Once we have placed the four consonants, there is only one way to arrange the four vowels in the remaining four positions so that they appear in alphabetical order.  Hence, there are
$$8 \cdot 7 \cdot 6 \cdot 5 = 1680$$
arrangements of the letters of the word AUROBIND in which the vowels appear in alphabetical order.
A: Another approach leading to the same result. Count how many different
positions vowels can occupy. It is just the number of combinations
$\binom{8}{4}$. Now for each combination there are 4! permutations of
the consonants. Therefore the total number is $\dfrac{8!4!}{4!4!}=1680.$
A: Just count the number of arrangements of all letters, $8!$, and divide it by the number of ways of ordering the vowels, $4!$ (because only one of the $24$ vowel orderings is valid). Easy! And it agrees with your second method.
