# Combination and permutation of indistinguishable objects

There is a group of 10 objects, 2 red, 3 blue and 5 green. The objects are indistinguishable. In how many ways can they be arranged on a line?

As there are 3 groups of objects I did that:

$10! / (2! * 3! * 5!)$

So my result is 25.200

Is that correct? I am reading a lot of material around, but I am still unsure.

thanks

• Your formula is correct but the calculated answer should be $2520$ not $25200$. – David Nov 12 '14 at 12:03
• uhm it still comes as 25200 to me? – jsab Nov 12 '14 at 12:15
• By my calculations it is indeed 2520. wolframalpha.com/input/?i=10!%2F%282!3!5!%29 – JMoravitz Nov 12 '14 at 12:19
• $$\def\\#1{\color{red}{#1}}\def\-#1{\color{green}{#1}}\frac{10!}{2!3!5!}=\frac{10.9.8.7.6.\\5.\\4.\\3.\\2.\\1}{2.1.3.2.1.\\5.\\4.\\3.\\2.\\1} =\frac{10.9.8.7.\-6}{2.1.\-3.\-2.\-1}=10.9.4.7=2520$$Just like playing around with colours, sorry ;-) – David Nov 12 '14 at 12:29
• I think it's preferable to add a few words to a question like this saying what kind of arrangement we mean. It appears you mean to arrange the objects in ten identifiable places, one object per place. If instead you arranged the objects symmetrically around a circle and considered two arrangements indistinguishable if one can be rotated to the other, you get a smaller answer. If you arrange the objects on an oriented chessboard, at most one object per square, you get more arrangements. – David K Nov 12 '14 at 13:04

To explain where your formula comes from and why it works, consider breaking it up into steps:

Step 1: choose where the reds go. There are $\binom{10}{2}$ ways of arranging the reds and not reds (ignoring the fact that the not reds are of multiple colors for the moment).

Step 2: of the spaces labeled for use by not-reds, choose which of those spaces will be occupied by blues: There are $\binom{10-2}{3}$ number of ways to do this.

Step 3: of the spaces labeled for use by not-reds and not-blues, choose which are occupied by greens: There are $\binom{10-2-3}{5}$ number of ways.

Thus, there are $\binom{10}{2}\cdot\binom{8}{3}\cdot\binom{5}{5} = \frac{10!~~~8!~~~5!}{2!8!3!5!5!0!} = \frac{10!}{2!3!5!}$ number of ways to accomplish this.

(remember that $0!=1$ by definition)

• thanks a lot for the explanation.. I have other 2 questions to solve.. I will open new questions and put my ideas to check them.. (or I am not sure if I should continue here..) – jsab Nov 12 '14 at 12:18
• It is best to open new posts. – JMoravitz Nov 12 '14 at 12:19