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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
user1172468
  • 527

3 Answers3

3

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 =$\frac{20!}{10!10!}$

Aang
  • 14,672
2

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.

André Nicolas
  • 507,029
2

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 $

hjpotter92
  • 3,049