Permutation and Combination-2

   In the previous post (Permutation and Combination-1), all four balls were different colors. However, in this post, we deal with the case that two balls are of the same color as shown in Fig. A-4. 

Fig. A-4 Two balls with the same color in four balls.

   In this case, how many arrangements can be distinguished by us if we line four balls up? By using the same method as Fig. A-3, we can obtain all possible ways to line four balls up as shown in Fig. A-5.

Fig. A-5 All possible 24 ways when two same colored balls are included.

   Although the indexing numbers of B1 and B2 are assigned in order to designate two blue balls, our eyes will perceive two blue balls as the same. In other words, we cannot distinguish each pair connected by arrows in Fig. A-5.
   Therefore, when four balls with two same color balls are lined up, we recognize that all possible ways to be distinguished are twelve as shown in Fig. A-6.

Fig. A-6 All possible 12 ways to be distinguished when two same colored balls are included.

   Furthermore, if three balls are the same color, we would say that all possible ways to be distinguished to us is only four as shown in Fig. A-7.

Fig. A-7 All possible 4 ways to be distinguished when three same colored balls are included.

   Finally, if all four balls are the same color, we know that there is only one distinguishable line up. Summarizing the discussion so far,

   (1) If four balls are all different colors and four balls are lined up in a row, all possible ways to be distinguished are 4!=24.

   (2) If two balls of four balls are the same color and four balls are lined up in a row, all possible ways to be distinguished are 24-12=12. We express this as twelve ways are redundancies or duplicated.

   (3) If three balls of four balls are the same color and four balls are lined up in a row, all possible ways to be distinguished are 24-20=4. We express this as twenty ways are redundancies or duplicated.

   (4) If four balls are the same color and four balls are lined up in a row, all possible way to be distinguished is only 1.

   If looking closely, we can see that the evaluated ways of (1), (2), (3) and (4) happen to be the same as 24=4!/0!, 12=4!/2!, 4=4!/3!, and 1=4!/4!, respectively.

   That is, the numerator's factorial corresponds to total number of balls and the denominator’s factorial corresponds to the number of balls with the same color. Therefore, when placing N balls with R same colored balls in a row, it can be easily inferred that all possible ways to be distinguished is simply N!/R!.

   Of course, there will be certainly any strict proof of N!/R!, but it may be sufficient that engineers simply use the formula of N!/R! leaving its proof to the mathematicians.