There are 'N' spaces and you have 'a' red balls and 'b' black balls (a+b = N). In how many ways can you place the balls in 'N' spaces so that the following condition holds: No two black balls are together. How would this result change if: Each red and black ball is distinct (a1, a2, b1 is a diff than a2, a1, b1) and we know the ball at the 'N'th posn has to be red.

A similar (although not identical question) was discussed earlier.

The answer to the first part should be (a+1 C b) and the answer to the second part should be (a+1 C b)* a!b! Is this correct?

QuoteOriginally posted by: sevvostA similar (although not identical question) was discussed earlier.So, do you have the correct answer to this question?

Balls indistinguishable:Place b black balls in a line. There are remain (a) red balls to be placed in (b+1) locations, of which the two at the end have >=0 red balls, and the other (b-1) in the middle have >=1 red balls.= partition (a+2) into at most (b+1) integers, each >=1.= partition (a+2 + b+1) into exactly (b+1) parts, each >=1So task is to Partition N into exactly R parts, P(N,R)The first part is >1 or =1, so we have:P(N+1,R) = P(N,R) + P(N,R-1) for N>R>1with initial P(x,x) = P(x,1) = 1 That's very basic now, so if that doesn't look familiar, look up gumdrop problem...

There needs to be at least 1 red ball between each pair of black balls. So the (b-1) spaces between the black balls have to be filled with red balls. That leaves us with a-(b-1) red balls to place. Now they can be placed either in the (b-1) spaces between the black balls, or in the 2 spaces at the ends. That can be done in C(a-b+1, b+1) ways. Quick check, when a=b-1, there is only one way to do it.