How many ways can we arrange 10 identical black balls and 10 identical red balls in a line, so that no ball has two neighbors that are of the same color?
Clarification: No ball (red or black) can be between two black balls or two red balls.
So, the following combinations aren't allowed for any three successive balls:
Image credit: https://www.dreamstime.com
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
I think it would be better to ask for the number of ways (assuming people of the same sex are identical), rather than the probability.
Log in to reply
But that would be a very large number. This is short and sweet.
Log in to reply
The number of ways is 4.
Note that I assumed "people of the same sex are identical". So yes, it might be better to rephrase it to "arrange coins showing 10 heads and 10 tails", or something like that.
Log in to reply
@Calvin Lin – OK @Calvin Lin I have updated the question. Can you update the answer to 4?
Log in to reply
@Geoff Pilling – Note, I used balls instead of coins to avoid any confusion involved with flipping of coins.
@Geoff Pilling – Perfect, thanks! I have updated the answer.
can you add that all 10 balls of same colour are identical?
What a about this RRRBBB....BBRRR..R. ???
Here a 3 is not different color from a 1 and also condition holds.
Log in to reply
What colors are the neighbors of ball 2?
Log in to reply
Both red which is allowable
Log in to reply
@Kushal Bose – The condition is "no ball has two neighbors that are of the same color"
What do you say "Both red which is allowable"?
Log in to reply
@Calvin Lin – Ohh I only consider RBR and BRB. Got it
Log in to reply
@Kushal Bose – Yea, we also cannot allow for RRR and BBB. Glad you pointed out this potentially common mistake :)
Log in to reply
@Calvin Lin – I have updated the wording for clarity.
Problem Loading...
Note Loading...
Set Loading...
Once you have placed the first two balls ( a 1 and a 2 ), the rest are determined as follows:
a n = A different color from a n − 2 for n > 2
There are two ways to arrange the first two balls:
So there are 4 ways all together.