Question

There are 12 balls of which 4 are red, 3 black and 5 white. In how many ways can you arrange the balls so that no two white balls may occupy consecutive positions if balls of the same colour are identical?

Answer 1

Hint: We need to find out the condition and rephrase it in such a way that the whole case can be divided in two parts. Then we arrange the non-conditional part for the red and black balls. We then find out the number of free spots to place the white balls for the conditional part.

Complete step-by-step answer:
The main condition in the arrangement of the balls is that no two white balls may occupy consecutive positions which means there has to be at least one red or black ball in between any two white balls.
So, in the arrangement we first arrange the red and black balls and then we arrange the white balls.
The only way white balls can be separated if we place them by only one at a time in between the gaps of any two red or black balls.
There are 4 red and 3 black balls. Balls of the same colour are identical. We first arrange these balls.
So, in total 7 balls and of two types. The arrangement of n things out of which r things are alike and the rest are alike, the arrangement becomes of the form $\dfrac{n!}{r!\times (n-r)!}$.
So, for our case the answer is $\dfrac{7!}{4!\times 3!}=35$.
Now we have distributed seven balls. We just need to place 5 white balls by only one at a time in between the gaps of any two red or black balls.
In those seven balls there are eight spots around them to place the white balls.
We just need to place 5 balls out of 8 spots. We arrange them in a combinational way.
We just need to choose 5 spots. This can be done in ${}^{8}{{c}_{5}}=\dfrac{8!}{5!\times \left( 8-5 \right)!}=\dfrac{8!}{5!\times 3!}=56$ ways.
So, the total number of arrangements is $56\times 35=1960$.

Note: We can not arrange the white until the red or black balls are distributed as they act as the dividers to keep the white balls separated. Any change in the arrangement of the unconditional part will create unnecessary arrangement for the conditional part. The spots can also be visualised using the concept of clock and the spots in between them.