# Satistics problem

#### MikeLin007

##### New Member
I have no idea where to start on this problem, could someone point me in the right general direction and I can try to work it out on my own. Here is the problem.

Say you have a strand of 206 beads in a random order. 5 of the beads are white and the rest of them are black. What is the probability that each of the white beads are separated by NO MORE than 50 beads. So if we cut the strand at each white bead, all of the cut strands will be NO MORE than 50 beads long.

And the same calculation but with 20 white beads and the rest black. Still 206 beads total.

Thanks!

#### jose

##### New Member
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:

#### MikeLin007

##### New Member
Thanks for your reply. Is there a way to find out how many different combinations are possible to satisfy the arrangement without having to list them all out and manually looking at them?

#### jose

##### New Member
Sorry that I was not clear enough. The formulas were all messed up when I posted them.

I only listed the arragements so you can see how it works.
You don´t have to list them, just apply the formula:

Possible arrangements = n! / (n1 !) . (n2 !)

(that is the factorial of n divided by the product of the factorial of n1 and n2). This will give you all the possible arrangements, One you get the result is simple because out of all the possible arragements, only one satisfies your condition.

In the example that i sent you