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# 10 figit numbers from 3 & 7

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by dsk » Wed Jul 09, 2008 2:16 pm
It would be similar to how can 2 numbers be arranged in 10 slots..the 1st digit of teh 10 digit number can be 3 or 7, so 2 ways. similarly the 2 nd digit can have 3 or 7, so two ways.. and so on..
so total arrangements would be 2*2*2...10 times = 2^10
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by VP_Jim » Wed Jul 09, 2008 3:34 pm
That's a nice explanation. I'm going to elaborate, to make this applicable to all permutation/combination problems:

I like to imagine these problems as "slots", like this:

__ __ __ __ __ __ __ __ __ __

Then, I just think: how many choices do I have for each slot? In this case, we have two choices (3 or 7) for every slot, so the answer is 2x2x2x2 etc., or 2^10.
Jim S. | GMAT Instructor | Veritas Prep
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