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

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The discussion centers on calculating the number of bit strings containing exactly eight 0s and 10 1s, with the condition that each 0 must be immediately followed by a 1. The initial assumption of 72 strings based on position calculations is corrected, as the 1s are indistinguishable. The correct approach involves treating the pairs of 0s and 1s as distinct units, leading to the arrangement of ten objects: eight pairs of (01) and two standalone 1s. The final calculation yields 45 unique bit strings, confirmed by the formula for combinations. The conclusion emphasizes the importance of recognizing indistinguishable elements in combinatorial problems.
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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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