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Given:

  1. A bag of 10 red balls
  2. A bag of 10 blue balls.

In how many sequences can one draw all the 20 balls.

so a sequence could be (r for red ball, b for blue ball):

b b b b b b b b b b r r r r r r r r r r

another sequence could be:

r b b b b b b b b b b r r r r r r r r r   
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1  
Is this homework? Should we be giving you the answer, or just hints? –  Jonathan Christensen Jan 25 '13 at 2:18
1  
Next step: What if there are 5 red and 15 blue? Or 7 red, 5 blue, 8 green? –  vonbrand Jan 25 '13 at 2:28
    
Hi, no not homework. Though I was thinking of opening the textbooks of my youth. How much my mind has decayed! –  user1172468 Jan 25 '13 at 3:35

3 Answers 3

up vote 1 down vote accepted

As you observed, the order of the draws can be represented by a sequence of length $20$ which has exactly $10$ $b$ and $10$ $r$. We can think of it as a word of length $20$, over the alphabet $\{b,r\}$, which has exactly $10$ $b$ and $10$ $r$. We want to count the number of such sequences (words).

A sequence is completely determined when we know the locations of the $10$ $r$. These $10$ locations can be chosen from the $20$ available in $\dbinom{20}{10}$ ways.

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You can see it as permutation of $20$ objects in which $10$ are of first kind(identical) and other $10$ are of second kind(identical).

Total number of such permutations =$\frac{20!}{10!10!}$

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Since you have total 20 objects to chose from, you can chose them in total of $ 20! $ ways.

But, the balls are all identical, so that decreases the number of possible cases.

Hence, you get the answer as $ C_{10}^{20} = \left( \matrix{20\\10} \right) = 184756 $

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