Possibly wrong permutation/combination question?

[stfz]

Homework Statement

In how many ways can 4 red balls, 3 white balls and 1 black ball be arranged in a line so that the black ball is always surrounded by a red and a white ball?

Textbook answer: 20

Homework Equations

Chapter was on permutations and combinations, so
$n P r = n!/(n-r)!$
$n C r = n!/(r!(n-r)!)$

The Attempt at a Solution

We have 4 red, 3 white, 1 black.
Black must be surrounded by 1 red and 1 white, so we treat RBW as one unit.
There are two possible combinations of this unit, RBW and WBR, so the result will be multiplied by 2.
There are 6 possible positions to place this 3 ball unit into an 8-ball space, so there are 6*2 = 12 different ways to place this unit.

For the remaining, we regard only order of color, so we use nCr:
For the red balls (only 3 remaining, and we have 5 spaces) there are 5 C 3 = 10 different ways to place them.
For the white balls (only 2 remaining, and we have 2 spaces), there are 2 C 2 = 1 ways to place them.

So the final result is : 12 * 10 * 1 = 120.

I would like a confirmation that this answer is correct (I'm usually not, but maybe I am just this once )

 

[Dick]

Homework Statement

In how many ways can 4 red balls, 3 white balls and 1 black ball be arranged in a line so that the black ball is always surrounded by a red and a white ball?

Textbook answer: 20

Homework Equations

Chapter was on permutations and combinations, so
$n P r = n!/(n-r)!$
$n C r = n!/(r!(n-r)!)$

The Attempt at a Solution

We have 4 red, 3 white, 1 black.
Black must be surrounded by 1 red and 1 white, so we treat RBW as one unit.
There are two possible combinations of this unit, RBW and WBR, so the result will be multiplied by 2.
There are 6 possible positions to place this 3 ball unit into an 8-ball space, so there are 6*2 = 12 different ways to place this unit.

For the remaining, we regard only order of color, so we use nCr:
For the red balls (only 3 remaining, and we have 5 spaces) there are 5 C 3 = 10 different ways to place them.
For the white balls (only 2 remaining, and we have 2 spaces), there are 2 C 2 = 1 ways to place them.

So the final result is : 12 * 10 * 1 = 120.

I would like a confirmation that this answer is correct (I'm usually not, but maybe I am just this once )

Sounds correct to me.

 

[Curious3141]

I get 120 as well. This is $\displaystyle \frac{6!}{3! 2! 1!} \times 2!$ which is another way of looking at it.

 

[mafagafo]

I also get 120.
It is twice a permutation of 6 elements (as I'm grouping WBR and RBW) with 3 repetitions of one type and 2 of another type.
$$2 \cdot P^{(3, 2)}_{6}=2 \cdot \frac{6!}{3!2!}=120$$