Can you give me a hint on how to approach b and c?

[Atria]

Do I take those 7 as 1 in a) and then arrange it with only the other 7 left?

 

[blamocur]

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Do I take those 7 as 1 in a) and then arrange it with only the other 7 left?

In b) there are no choices for the first 7, so it is one possibility, but then you have multiple possibilities for the remaining 7 questions. So, how many possibilities are there to answer 7 questions so that at least 1 of them is correct?

 

[Atria]

In b) there are no choices for the first 7, so it is one possibility, but then you have multiple possibilities for the remaining 7 questions. So, how many possibilities are there to answer 7 questions so that at least 1 of them is correct?

okay got b) so I just had to calculate C(7,1)+C(7,2)+...+C(7,6)+C(7,7) = 127

What about c)?

 

[Steven G]

What about c? I am not going to spend my free time trying to figure out what you dod to get your answer--that's your job. Please inform us how you got your answer.

 

[Atria]

okay so it is C(7,6)*120+C(7,7)*127 = 967

 

[Steven G]

I still question b and c.
You claim that there are 127 possibilities for part b. Can you please list 10 of these 127?

Edit: b and c correct, sorry about that.

 

[Dr.Peterson]

okay got b) so I just had to calculate C(7,1)+C(7,2)+...+C(7,6)+C(7,7) = 127

What about c)?

That can be done a little more easily: There are 2^7 = 128 ways to answer the last 7 questions, only one of which has all of them wrong, so there are 128-1 = 127 ways to get at least one right.

okay so it is C(7,6)*120+C(7,7)*127 = 967

That's correct, though it would be more helpful if you stated the reason for each number, in case your thinking is wrong. The 127 is as above, and the 120 (the number of ways to get at least 2 right of the last 7) can be obtained the same way, as "all ways - 0 right - 1 right" = 2^7 - C(7,0) - C(7,1) = 128 - 1 - 7 = 120. Is that what you did?

[I was locked out for an hour, even from getting in on a different computer with a different browser without signing in ...]

 

[Atria]

That can be done a little more easily: There are 2^7 = 128 ways to answer the last 7 questions, only one of which has all of them wrong, so there are 128-1 = 127 ways to get at least one right.

That's correct, though it would be more helpful if you stated the reason for each number, in case your thinking is wrong. The 127 is as above, and the 120 (the number of ways to get at least 2 right of the last 7) can be obtained the same way, as "all ways - 0 right - 1 right" = 2^7 - C(7,0) - C(7,1) = 128 - 1 - 7 = 120. Is that what you did?

[I was locked out for an hour, even from getting in on a different computer with a different browser without signing in ...]

Yes, exactly. I discounted 7 from my previous answer instead

I think C(7,0) is unnecessary

 

[Dr.Peterson]

Yes, exactly. I discounted 7 from my previous answer instead

I think C(7,0) is unnecessary

It isn't unnecessary; you used it! You first subtracted that 1 from 128 to get 127, and then subtracted the 7 to get 120. I just did them both at once, starting from the 128.

 

[Atria]

Okay, I see what you mean. you basically counted all the variants and discounted from that the 2 variants that I do not need, is that correct?

 

[Atria]

In b) there are no choices for the first 7, so it is one possibility, but then you have multiple possibilities for the remaining 7 questions. So, how many possibilities are there to answer 7 questions so that at least 1 of them is correct?

I still question b and c.
You claim that there are 127 possibilities for part b. Can you please list 10 of these 127?

Edit: b and c correct, sorry about that.

It isn't unnecessary; you used it! You first subtracted that 1 from 128 to get 127, and then subtracted the 7 to get 120. I just did them both at once, starting from the 128.

thanks for the help