Number of ways of arranging 7 characters in 7 spaces

[Aurelius120]

Homework Statement

1) What are the number of different ways of arranging $2,2,2,3,x,y,z$ in seven spaces $—,—,—,—,—,—,—$ such that the rightmost position always has a letter?
$$OR$$
2) What is the number of ways of distributing $2,2,2,3$ in three boxes, $x,y,z$ such that every number is contained in a box?

Relevant Equations

NA

The rightmost position has 3 possibilities: $x,y,z$
The remaining two letters are to be arranged in 6 spaces: $\frac{6!}{4!}$
Now the 3 can be placed in $\frac{4!}{3!}$
Total no of ways =$$3×\frac{6!}{3!}=12×30$$
$$OR$$
Since $x,y,z$ are three different boxes/variables, we can use the Stars and Bars method which gives: 7 characters in 7 spaces with a letter at rightmost position if every number to the left of a letter is assumed to be contained in the box and the solution is as above.

Am I correct?
I don't see how this is different from finding three non-negative integers (x,y,z) such that they are the solutions of $xyz=24=2×2×2×3$ The answer to this however is 30.


So what am I missing? Why is the 'Stars & Bars' method not working for question in blue? How are the 3 questions different if at all?

EDIT1:
Since x,y,z are unique variable answer by the Star Bar method of the third question should be similar.(Maybe half as pointed by @Hill ; certainly not 12 times less)

 

[Hill]

How are the 3 questions different if at all?

They are different because when you put 2 and 3 or 3 and 2 in two boxes, they are two different ways, but 2x3 and 3x2 are one solution in the third question.

 

[BvU]

everything to the left of a letter is assumed to be contained in the box

you mean 'every number', not: 'everything' , am I right ?
And two adjacent letters are not allowed -- right ?

$\$

 

[Aurelius120]

you mean 'every number', not: 'everything' , am I right ?
And two adjacent letters are not allowed -- right ?

$\$

Yes sorry every number. Two adjacent letters should make one box empty which is allowed in distributing the numbers

 

[Aurelius120]

Ok Now I think, I am getting it
The sequence of variables $(x,y,z),(x,z,y)$ OR whatever
Only different values of the fixed sequence $(x,y,z)$ matter
But how to solve the third question in blue?
I tried this way:
There are six spaces: $—,—,—,—,—,—$
There are two separators: $|,|$ and four numbers that have to be placed: $2,2,2,3$
Number of ways= (Arrangements of Separators)×(Arrangements of Numbers)$$=\frac{6!}{2!.4!}\times \frac{4!}{3!.1!}=60=2×30$$
This is double the correct value.
This is obviously wrong.
How to do this correctly using the Stars and Bars Method?

In General:
What then is the correct way of arranging, say $m$ balls of $m_1,m_2,....m_n$ number of balls of $n$ types in $r$ boxes using Stars and Bars method?

 

[Hill]

Maybe half as pointed by @Hill ; certainly not 12 times less

I think it makes it 12 times less: there are 12 different ways to put 2 and 3 in 4 boxes and they all are the same for the third question.

 

[Hill]

There are two separators: |,| and four numbers that have to be placed: 2,2,2,3
Number of ways= (Arrangements of Separators)×(Arrangements of Numbers)

On one hand, you double count, e.g., you get solutions like 2x(2x3)x2 and 2x(3x2)x2, which are the same 2x6x2.
On the other hand, isn't 1 an allowed value for x,y,z?

 

[Aurelius120]

On one hand, you double count, e.g., you get solutions like 2x(2x3)x2 and 2x(3x2)x2, which are the same 2x6x2.

Oh that's why it's giving double the correct answer.

On the other hand, isn't 1 an allowed value for x,y,z?

Yes that's what happens when two $|$ are adjacent or variable is empty of $2,3$

 

[WWGD]

Haven't read in full detail, but I think the Multinomial Coefficient may apply here.