Help with Discrete Math I "Counting" Homework

[theADOLESCENT]

Hey guys, we just started learning "counting" and I have a few questions about my homework.

 

1) How many 5-digit integers start and end with an odd digit?

 

I figure 99999 is the highest numbers so it would go like this.

 

(5)(9)(9)(9)(5)

because for the first number you have 5 choices (1,3,5,7,9), the 2nd,3rd,4th numbers have 10 choices (0,1,2,3,4,5,6,7,8,9)
and the last has the same amount of choices as the first number. So I got 25,000 as my final answer. That one wasn't hard.

 

2)How many 5-digit integers begin with an odd digit, end with an odd digit, and consist of 5 different integers?

Again, 99999 would be highest number possible if they didn't have to be all different.

 

Here's what I get: (5)(9)(8)(7)(4 or 1?) because for first choice you have 5 choices (1,3,5,7,9), second you have (0,1,2,3,4,5,6,7,8,9)-your first choice so 9 choices, next you have minus first and 2nd choice so 8 choices, 3rd you have minus 1st,2nd, and 3rd so that's 7 choices, but I'm confused for the last one. If the first 4 choices are all odd, then you're left with 1 choice for the last number, if the first number is odd, but the 2nd, 3rd, and 4th are even, then you have 4 choices.

I'm confused about that one...

 

3) How many 5-digit integers start with an odd digit, end with an odd digit, and contain exactly two zeroes.

 

I did: (5)(9)(1)(1)(5) and got 225. I'm not 100% sure about this one...

 

Any help is appreciated.

 

Thanks,

theADOLESCENT

[John]

For number 2, start with the greatest restriction, i.e. the first and last digits must be odd. But remember that the first and last digits can't be equal. Once those are chosen, then look at the middle three digits, keeping in mind that they can't match each other and they also can't match the first or last digit.

 

For number 3, remember that there are multiple ways to have exactly two zeroes. Your answer only accounts for one of those ways. For instance (though this may be giving too much away), your answer includes the number 34005, but not 30405 or 30045.

Edited December 4, 2012 by John

[theADOLESCENT]

Thanks,

 

I understand how my #3 doesn't account for those, it's just the figuring out where to go from there haha.

 

Would I do something like (5)(9)(1)(1)(5) + (5)(1)(9)(1)(5) + (5)(1)(1)(9)(5) = 675?

 

 

My redone #2 is (5)(8)(7)(6)(4) with doing it your way for a product of 6,720.

[John]

Correct on both counts.

 

Just to be sure I'm not insane, I also wrote a couple of small programs to find both answers manually, and they agree.

 

I don't know if you've covered combinations yet, but the reason we multiply by three is, more formally, we want to find the number of ways the middle three digits can include exactly two zeroes, which is equal to [math]9\times{{3}\choose{2}}[/math] (the parenthetical part is read "3 choose 2" and equals [math]\frac{3!}{2!(3-2)!} = 3[/math]). So we choose the first and last digits from the odd digits, giving us [math]5\times5[/math], then multiply that by the [math]9\times{{3}\choose{2}}[/math] mentioned before to arrive at [math]5\times5\times9\times3 = 675[/math].

Edited December 4, 2012 by John

[theADOLESCENT]

Thanks a lot.

I don't really get the stuff you're talking about near the end up looking ahead in the notes it looks like we could be learning something along those lines.

 

theADOLESCENT

[John]

You're welcome.

 

Don't sweat that last part. Given the kinds of problems you're apparently working on now, it'll probably be covered before too long, and you'll probably find it's a pretty handy operation.

[Izzy_Bee]

Thanks a lot.

 

I don't really get the stuff you're talking about near the end up looking ahead in the notes it looks like we could be learning something along those lines.

 

theADOLESCENT

You seem to have taken to this well and understood it pretty well too, so I am sure in the end you will understand EVERYTHING John had to say, you'll have to excuse some of us...we get...carried away when we see something that we enjoy come up.

But, I have faith in you and I am sure you've not far to go at all! Hope you do well and pass what you need to pass!