Question

6 red and 6 black balls are arranged such that no two black balls are together. Find the total number of ways it can be arranged.

Answer 1

The number of ways to arrange the black balls without restrictions is 6! (factorial), as there are 6 black balls.
Now, let's consider the red balls. Since there are 6 red balls and 7 available slots, we can select 6 slots out of 7 for the red balls. This can be calculated using combinations, denoted as "7 choose 6" or C(7, 6), which equals 7.To obtain the total number of arrangements, we need to multiply the number of arrangements of the black balls by the number of arrangements of the red balls:
Total number of arrangements = Number of arrangements of black balls * Number of arrangements of red balls
= 6! $\times$ C(7, 6)
= 6! $\times$ 7
= 720 $\times$ 7
= 5040
Therefore, the total number of ways to arrange 6 red and 6 black balls such that no two black balls are together is 5040.