In how many ways we can order 20 balls in line with non-adjacent block rule of 5 red balls

Question

We have 20 balls in two colors and two different sizes.

5 big red (R)

5 small red (r)

5 big white (W)

5 small white (w)

In how many ways we can order them in line with this rule:

Red balls must be in blocks of 5 balls and cant be together (adjacent) with other block of 5 red balls.

Note: Balls with same size and same color we dont distinguish.

example of this kind of order: (W-w-R-R-r-R-r-w-w-w-R-R-r-r-r-W-W-W-w-W)

Answer 1

You can count the possible placements of the blocks, and then place the balls accordingly.

Label the positions in the list with the interval $[1,20]$, and let $b_1$ and $b_2$ be the positions of the first ball in each block. The first block can start at positions $b_1\in[1,10]$, and the second block can start at positions $b_2\in[b_1+6,16]$. Therefore the total number of distinct placements of blocks is given by:

$$\sum_{b_1=1}^{10}\ \sum_{b_2=b_1+6}^{16}1=\sum_{b_1=1}^{10}(11-b_1)=\sum_{b_1=1}^{10}b_1=55$$

For each of these positions, there are $10$ positions to be filled by red balls and $10$ to be filled by white balls. Because each of these $55$ block placements indicates a specific arrangement of red and white, irrespective of size, the red and white balls can be placed independently. This adds a factor of $\binom{10}5$ for each color. The total is thus:

$$55\binom{10}5\binom{10}5=349720$$

Answer 2

Hint:

• How many ways are there to arrange 5 $w$s and 5 $W$s?

• For each of these arrangements, we can choose two dividers to place into the arrangement, representing the two blocks of five red balls. There are $11$ positions for the dividers, so $\binom{11}{2}$ ways to place the dividers.

• Finally, how many ways can we arrange the ten red balls into two blocks of five? This should just be the same as the number of ways to arrange 5 $rs$ and 5 $R$s. Do you see why?