The ten digits 0,1,2,3,...9, are arranged at random with no repeats. What is the probability that the number thus formed represents
a.)a # greater than 6 billion?
b.) an even # greater than 6 billion?(HINT- There are two cases to consider: 1st digit odd and 1st digit even)

Question

kinggeorge · Answer

Well, for a, first we have to find how many numbers above 6 billion we can make. As it turns out, if we have a first digit of 6 or greater, the number will be greater than 6 billion. So the first digit can be 6, 7, 8, or 9. That leaves us with 9 other digits.

This is a time for permutations. We have 9 spaces, and 9 digits, so have $9!$ ways to represent the last 9 digits. We also have 4 possible numbers for the first digit, so we have $$4\cdot9!$$ways to make a number greater than 6 billion.

kinggeorge · Answer

For b, it's more complex. 
Case 1: Even digit in front
            This means that you either have a 6, or an 8 in front. Either way, you have 4 other even digits left. Once of these must go at the end. Then you have 8 digits left to fill in the rest of the places. So your answer should be $$2\cdot8!\cdot4$$Two choices for the first digit, 8! choices for the middle 8 digits, and 4 choices for the last digit.

Case 2: Odd digit in front
            Now you have a 7 or a 9 in the front. This means that you have 5 even digits to choose from to use for the last digit. So your answer would be $$2\cdot8!\cdot5$$Two choices for the first digit, 8! choices for the middle 8 digits, and 5 choices for the last digit.

anonymous · Answer

OH! I see, so this is like those problems where you put all those blanks and fill them with numbers like 4.9.8.7.6.5.4.3.2.1? (The periods are the multiplication signs)

kinggeorge · Answer

That's a good way to think about it.

anonymous · Answer

ya, I like it when it's a bit more visual so I can see how things work out.  So what happens  next to the 2.8!.5 and 2.8!.4?

kinggeorge · Answer

What do you mean by "happens next to ..."?

anonymous · Answer

Do you do anything else with them, or is that the end of the problem?

anonymous · Answer

When they become fractions for the probablity, how do you do that?

kinggeorge · Answer

Well, since it's asking for the total probability of the number being odd, you would have to add them together next. So you get $$(2\cdot8!\cdot4)+(2\cdot8!\cdot5)=2\cdot8!\cdot(4+5)=2\cdot8!\cdot9=2\cdot9!$$Then to find the probability, divide this number by the total amount of numbers above 6 billion.$${2\cdot9! \over 4\cdot9!}={1 \over 2}$$

anonymous · Answer

The answer sheet says that it is 1 over 5... hmmm... very peculiar...

kinggeorge · Answer

Ah. I see what it was asking for the probability of now. Instead of dividing by $4\cdot9!$ we need to divide by the total amount of numbers you can create. Since you have 10 digits, and 10 places, you can make $10!$ total numbers. So the probability is $${2\cdot9! \over 10\cdot9!}={2 \over 10}={1 \over 5}$$

kinggeorge · Answer

Likewise, for a, you want $${4\cdot9! \over10\cdot9!}={2\over5}$$

anonymous · Answer

Thank you so very much KingGeorge. You are a god! 

I know that this will help me tons on the math quiz I have on this tomorrow!!! Wish me lots of luck!!!Actually, that's an insult to math people.Wish me lots of ...skill and... logic!!!

kinggeorge · Answer

Good skill and logic! With maybe just a little bit of luck thrown in there for good measure.

anonymous · Answer

hey can anyone help me too with functions???...:(
i'm having a problem in it...

anonymous · Answer

Haha! Thank you so much! I appreciate it more than you would know!!! Bye, I got to go to sleep so I can ace that quiz tomorrow! Good skill and Logic to you too!!!

kinggeorge · Answer

Good night.