1. ## combinations

7. If a year has 364 days, then the same calendar could be used every year by only changing the year. A "regular" year has 365 days and a leap year has 366 days. The year 2000 has a leap year and leap years occur every 4 years between the years 2000 and 2100. Claudia has calendar for 2009. What will be the next year that she can use this calendar by merely changing the year?

8. A 6-question True-False test has True as the correct answer for at least 2/3 of the questions. How many different true/False answer patterns are possible on an answer key for this test?

2. 7. If a year has 364 days, then the same calendar could be used every year by only changing the year. A "regular" year has 365 days and a leap year has 366 days. The year 2000 has a leap year and leap years occur every 4 years between the years 2000 and 2100. Claudia has calendar for 2009. What will be the next year that she can use this calendar by merely changing the year?
The reason calendars can't be reused is because the day of the week of a date changes. Suppose that the first day of January in 2009 is a Thursday. Then 365 days later, in 2010, it will be first of January on a Friday, because 364 is a multiple of 7 and so the day of the week increases by 365-364=1.
If you get another first of January on a Thursday, and it isn't a leap year that year, then the calendar can be reused.

8. A 6-question True-False test has True as the correct answer for at least 2/3 of the questions. How many different true/False answer patterns are possible on an answer key for this test?
2/3 of 6 =4 so there must be at least 4 Ts. There is no easier way to handle this than adding up the numbers of possibilities with 4, 5 and 6 Ts.

3. 7. I thought the answer was 2010, I just needed to make sure.

8. As far as this one you're telling me that I should add up 4! +5! + 6! ???
or can I do something like 6! = 720 - 2/3???

or are you saying 3! here cause you have to have at least 4 true answeres no matter the pattern??

4. 7. I thought the answer was 2010, I just needed to make sure.
It isn't.

8. As far as this one you're telling me that I should add up 4! +5! + 6! ???
or can I do something like 6! = 720 - 2/3???
The number of possibilities with exactly t Ts in n questions is given by ${n \choose t} = \frac{n!}{t!(n-t)!}$

5. 7. Okay now I understand it would be 7 years from 2009 but gotta figure in the 1 leap year that would be in that 7 year period so it would be 6 years from 2009 which would make it 2015??

8. I'm still not getting #8 sorry.

6. ## Still Lose In How To Solve #8

I guess I'm just not getting #8.

7. The number of possibilities with exactly t Ts in n questions is given by
The number of possibilities with exactly 4 Ts in 6 questions is ${6\choose 4}$.
The number of possibilities with exactly 5 Ts in 6 questions is ${6\choose 5}$.
The number of possibilities with exactly 6 Ts in 6 questions is ${6\choose 6}$.

So the number of possibilities with at least 4 Ts is
${6\choose 4}+{6\choose 5}+{6\choose 6}$

8. Now when I solve i come up with a fraction. 6/4 + 6/5 + 6/6 = 252/60

I know youve been very helpful here Bad, but I guess i'm just not getting this one.

9. Now for like 4 t's i get this

n/t = n!/t!(n-t)!

6/4 = 6!/4!(6-4)!
6/4 = 6*5*4*3*2*1/4*3*2*1(2!
6/4 = 6*5/2*1
6/4 = 30/2
6/4 = 15
6=60
= 10 for 4t's

10. n/t = n!/t!(n-t)!

6/4 = 6!/4!(6-4)!
6/4 = 6*5*4*3*2*1/4*3*2*1(2!
6/4 = 6*5/2*1
6/4 = 30/2
6/4 = 15
You should have stopped here.

${6\choose 4} = 15$
Also ${6\choose 5} = 6/1! = 6$ and ${6\choose 6} = 1/0! = 1$

11. Thank You so much Bad for steering me in the right direction instead of just driving me there. Cause of the way you helped me here I learned what to do instead of just getting the answer!!