- #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.

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.

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)

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)

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