**Probability problem**

In order to solve the problem, what yo need to know is in how many different ways you can arrange the 206 beads. So, what what you have in hands is a permutation exercise in which some of the objetcs are identical.

The equation to find out how many different arrangements you can have (when some objetcs are identical, is the following:

Permutation (when same things are identical) = n!

(n1!) (n2!)

I´ll make an exemple using just 5 beads (3 of them being white and 2 of them being black)

n: 5

n1: 3

n2: 2

P = 5! = 5 x 4 x 3 x 2 x 1 = 120 = 10

(3!) (2!) (3 x 2 x 1) . (2 x 1) (6). (2)

So, you can have 10 diffente arregements, and only 1 of them (arrangement 7) satisfies the requirement of having 3 white beads between 2 black ones, as shown below.

Arrangement o 5 beads (3 Whites (W) and 2 Blacks (B))

1. B B W W W

2. W B B W W

3. W W B B W

4. W W W B B

5. B W B W W

6. B W W B W

7. B W W W B

8. W W B W B

9. W B W B W

10 W B W W B

Finally the probability is 1/10

Hope it helps! :wave: