I'm have a little problem grasping string concepts. Here's my other question. Please provide explanation. How many 8 bit strings contain exactly three 0’s?

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- anonymous

is 8 bit 8 binary digits?

- anonymous

yes

- anonymous

so there are 32 different possible combinations right?

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- anonymous

I think so yes.

- anonymous

Not 32 what was I thinking its 2^8 which is 256

- anonymous

can you explain please?

- anonymous

8 binary digits, each binary digit can be either 1 or 0, so I did 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 or 2^8

- anonymous

but the question ask how many 8 bit string contain EXACTLY three 0's?

- anonymous

right, but I'm just saying there are 256 possible combinations

- anonymous

I don't understand the concept. what if it asked contain EXACTLY four 0's would the answer be the same?

- anonymous

No no no no , 256 is not the answer. I'm just saying there are 256 DIFFERENT combinations for binary in 8 digits

- anonymous

OK so how do I figure out the answer?

- anonymous

Now to find the answer we can do 8 choose 3, it basically means how many combinations of 3 things can come out of 8 places for lack of better words. The formula for this is (for a choose b)
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- anonymous

a!/b! x (a-b)!

- anonymous

so try to figure it out now.

- anonymous

ok thank you

- anonymous

the answer is 56 i believe.

- anonymous

I don't under what (" for a choose b") mean. Can you explain please.

- anonymous

I just meant that when I say "a choose b" where a is bigger than b I mean the formula that I showed below

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