Finding the Number of Bit Strings with 8 Zeros and 10 Ones

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The discussion focuses on calculating the number of bit strings containing exactly eight 0s and ten 1s, with the condition that each 0 must be immediately followed by a 1. The correct approach involves treating each pair of 0 and 1 as a single unit (01), resulting in the arrangement of ten objects: (01),(01),(01),(01),(01),(01),(01),(01),1,1. The final calculation yields 45 distinct bit strings, derived from the formula $$\dfrac{10!}{8!2!}$$, confirming that the indistinguishability of the 1s is crucial to the solution.

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How many bit strings contain exactly eight 0s and 10 1s if every 0 must be immediately followed by a 1?
 
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Alexmahone said:
How many bit strings contain exactly eight 0s and 10 1s if every 0 must be immediately followed by a 1?

Attempt? (Wondering)
 
I like Serena said:
Attempt? (Wondering)

x01x01x01x01x01x01x01x01x

I thought that since there are 9 positions for the 9th 1 and 8 positions for the 10th 1, there would be 9x8=72 strings. But this is just wrong since the 1s are indistinguishable.
 
Alexmahone said:
x01x01x01x01x01x01x01x01x

I thought that since there are 9 positions for the 9th 1 and 8 positions for the 10th 1, there would be 9x8=72 strings. But this is just wrong since the 1s are indistinguishable.

Two scenarios: either one of those x'es is 11 and the others are empty, or 2 of those 9 x'es are 1. (Thinking)
 
My thinking goes like this.

You have ten objects to arrange: (01),(01),(01),(01),(01),(01),(01),(01),1,1

So the answer is $$\dfrac{10!}{8!2!}=45$$
 

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