# Math Help - Ordering Probability

1. ## Ordering Probability

Five red balls, two white balls and three blue balls are ordered randomly on a table. Suppose balls of the same color can not be distinguished
from each other (i.e. the ordering of balls with the same color does not
matter). What is the probability that balls with the same color are ordered
together (i.e. there are no blue or white balls between red balls, no red or
blue balls between white balls, and no red or white balls between blue balls)
?

2. Originally Posted by billy
Five red balls, two white balls and three blue balls are ordered randomly on a table. Suppose balls of the same color can not be distinguished
from each other (i.e. the ordering of balls with the same color does not
matter). What is the probability that balls with the same color are ordered
together (i.e. there are no blue or white balls between red balls, no red or
blue balls between white balls, and no red or white balls between blue balls)
?
There are alternative ways to examine this.
Most simply is to take all the reds as "Group R", the whites as "Group W", the blues as "Group B".

Then the probability is the number of arrangements of the groups divided by the number of arrangements of the balls if you numbered them with labels from 1 to 10. That way you can see the different arrangements irrespective of their colours.

If we numbered the balls we can calculate the number of arrangements
with the balls together side by side in their own groups as

$\frac{5!3!2!\color{blue}(3!)}{10!}=\frac{3(2)2(3)2 }{10(9)8(7)6}=\frac{1}{7(5)4(3)}$

We multiply by 3! to arrange the groups around each other.

In any arrangement, the 5 reds can be arranged in 5! ways.
We cannot distinguish them if the balls are not numbered or differ in some way. So we divide the total number of arrangements by 5! to find the number of arrangements we can't distinguish due to the reds looking the same. We do the same for the whites and the same for the blues.

Hence we must divide by 5!, 3! and 2!

Therefore, the number of "indistinguishable" arrangements in total is

$\frac{10!}{5!2!3!}=\frac{10(9)8(7)6}{2(3)2}=5(9)8( 7)$

Hence the probability that the balls line up in their respective groups is

$\frac{3!}{5(9)8(7)}=\frac{1}{7(5)4(3)}$

3. Helolo, billy!

Edit: corrected a silly error.
Thanks for the heads-up. Archie!

Five red balls, two white balls and three blue balls are ordered randomly on a table.
Suppose balls of the same color can not be distinguished from each other.

What is the probability that balls with the same color are ordered together?

There are: . ${10\choose5,2,3} \:=\:\frac{10!}{5!\,2!\,3!} \:=\:{\color{blue}2520}$ possible arrangements.

If the same colors are adjacent, there are $3! = 6$ arrangements:

. . $\{RRRRR|WW|BBB\},\;\{RRRRR|BBB|WW\}. \;\{WW|RRRRR|BBB\}\;\text{ . . . etc.}$

Therefore: . $P(\text{adjacent colors}) \;=\;\frac{6}{2520} \;=\;{\color{blue}\frac{1}{420}}$