Hard Perms and Combs Questions Need Help~ (1 Viewer)

[Astraea]

i) A binary string is a sequence of 1s and 0s E.g. 110111100101 is a binary string of length 12. In a binary string of length 50, how many ways are there to have a string with exactly 9 1s and no two 1s are adjacent?

ii) Given 50 cards with integers 1,2,3...50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

 

[braintic]

(i) I believe the answer is 40C9.
Place the 41 zeros in a line. There are 40 spaces between the zeros. Then pick 9 of the 40 spaces to places the 1s in.
I believe that covers all the cases and nothing but the cases. But I'm prepared to be corrected.

 

[HSC2014]

Why do you only consider the spaces between the zeroes? Can't 1 be the first/last digit?

 

[braintic]

Why do you only consider the spaces between the zeroes? Can't 1 be the first/last digit?

Of course. So 42C9. Does that seem correct?

 

[braintic]

ii) Given 50 cards with integers 1,2,3...50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

Just seeking a clarification with Q2 (not that it will help me as yet):

Do you want ordered or unordered selections?

 

[HSC2014]

Bump! Curious for solution to ii)

 

[braintic]

ii) Given 50 cards with integers 1,2,3...50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

Do you have the answer to no. (ii) ?

Assuming you want unordered selections, just wondering if it works out to be 40C9 + 2(40C8) + 40C7 ?

 

[Astraea]

Do you have the answer to no. (ii) ?

Assuming you want unordered selections, just wondering if it works out to be 40C9 + 2(40C8) + 40C7 ?

Umm,i don't atm. But I can get it 2moro. Its hw from coaching. Btw, can you explain (i) and (ii) in detail because Perms and Combs really messes with my head haha. Thanks.

 

[braintic]

Umm,i don't atm. But I can get it 2moro. Its hw from coaching. Btw, can you explain (i) and (ii) in detail because Perms and Combs really messes with my head haha. Thanks.

For (i), all I can do is repeat my earlier post with the correction for my error:

I believe the answer is 42C9.
Place the 41 zeros in a line. There are 42 places where you can place the 1s, that is in the 40 spaces between the zeros, and 2 more places at the ends of the line.
Then pick 9 of these 42 spaces to place the 1s in.
There will be no more than a single 1 between each of the zeros.

For (ii), I am nowhere near confident, and there is not much point describing how I got this until I know if it is correct.
I'm not even sure I could explain it well anyway. But if you need to hand it in for tutoring homework, I'll give it a go tomorrow morning - just let me know.

I have to say that these questions are extremely hard even for Ext 2, and I'm sure that (ii) would never be asked in an HSC exam (unless there is a much easier way of doing it that I haven't been able to see).

 

[Astraea]

For (i), all I can do is repeat my earlier post with the correction for my error:

I believe the answer is 42C9.
Place the 41 zeros in a line. There are 42 places where you can place the 1s, that is in the 40 spaces between the zeros, and 2 more places at the ends of the line.
Then pick 9 of these 42 spaces to place the 1s in.
There will be no more than a single 1 between each of the zeros.

For (ii), I am nowhere near confident, and there is not much point describing how I got this until I know if it is correct.
I'm not even sure I could explain it well anyway. But if you need to hand it in for tutoring homework, I'll give it a go tomorrow morning - just let me know.

I have to say that these questions are extremely hard even for Ext 2, and I'm sure that (ii) would never be asked in an HSC exam (unless there is a much easier way of doing it that I haven't been able to see).

Umm, my class starts 2:00pm. I can probably get the answer for the questions around 11ish since there's a homework help class where the Tutors help the students out. Haha, well my tutoring is renowned for maths, I guess the streak must continue, hence hard questions. Thanks for the help btw, its up to you if you want to give it a go, I'll just tell you if its right or wrong at the end of the day.
P.S: Do you reckon you can help me with other questions of equal difficulty if I post it up?

 

[v1]

i) A binary string is a sequence of 1s and 0s E.g. 110111100101 is a binary string of length 12. In a binary string of length 50, how many ways are there to have a string with exactly 9 1s and no two 1s are adjacent?

ii) Given 50 cards with integers 1,2,3...50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

iirc thats from 2012 SBHS 4u trials or maybe 3u i forgot

 

[braintic]

I'm starting to get more confident that my earlier answer to (ii) is correct as it is numerically equal to my answer using a different method.

My previous answer was 40C9 + 2(40C8) + 40C7.
This is the same as 42C9.

The logic is as follows:

The number of ways to pick 9 numbers from the first 42 integers is 42C9. If we arrange them in increasing order and add 1 to the 2nd number, 2 to the 3rd number, 3 to the 4th number, ... 8 to the 9th number, then we get a set of numbers ranging from 1 to 50 which must be separated by at least 2, which is what is required.

Following that logic, I'm guessing that if we instead required each pair of numbers to be separated by at least 3 (ie. at least two integers between them) then there would be 34C9 possibilities.
This is consistent with expectations as we increase the gap. If we require 5 numbers between any pair we get 10C9=10. If we think about this, a possible combination is 1, 7, 13, 19, 25, 31, 37, 43, 49. The only way of getting other combinations that work is to increase each of the numbers one at a time starting from the last number, giving 10 possibilities.

And if we now increase the size of the gap to 6, we get 2C9, which is meaningless as expected.

 

[cineti970128]

i) 42C9
ii) 42C9 !!!!

 

[braintic]

i) 42C9
ii) 42C9 !!!!

Thanks for repeating the answers.