Number of ways of arranging 7 characters in 7 spaces

  • #1
Aurelius120
168
17
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.
20240129_083643.jpg


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)
 
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  • #2
Aurelius120 said:
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.
 
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  • #3
Aurelius120 said:
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 ?

##\ ##
 
  • #4
BvU said:
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
 
  • #5
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?
 
  • #6
Aurelius120 said:
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.
 
  • #7
Aurelius120 said:
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?
 
  • #8
Hill said:
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.
Hill said:
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##
 
  • #9
Haven't read in full detail, but I think the Multinomial Coefficient may apply here.
 

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