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In a bag there are 10 indistinguishable balls. Four of them are white and $6$ are black. The balls are taken out one by one and put on the table as they are taken.

In how many ways we can get at least $2$ consecutive white balls?

When I got this problem, I immediately thought about the opposite event. Let $A \space$ be "get at least $2$ consecutive white balls". The opposite event will be "don't get any consecutive white balls"

This will mean that the white balls have to be between the black balls or in the ends of the sequence.There are $6$ black balls, so:

_B_B_B_B_B_B_

There are $7$ positions that the $4$ white balls can occupy. My doubt is: It's correct one "invented" $3$ more positions in the sequence of $10$ balls? The assumption was if the positions were not occupy by the white balls, then they simply desappear.

If this thought is correct the final step is to find what is been ask.

The total of ways (without restrictions) is given by: $\frac{10!}{6! \cdot 4!}$

Then what is wanted is: $\frac{10!}{6! \cdot 4!}- {}^7C_{4}$

Thanks

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up vote 1 down vote accepted

You're right on track with this! Nicely done.

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