S1-Paper 62. Help please. ASAP.

[shah zaib]

I did the weirdest thing for the Year group question
I did 7C1 x 2C1 x 2C1 (for one of each) x (6C2 + 6C1 x 1C1 + 6C1 x 1C1 + 1C1 x 1C1) (for the other three)
Which is 28 x 28
784

i did exactly the same thing. And i believe this shouldnt be wrong. :/

 

[randomcod]

i did exactly the same thing. And i believe this shouldnt be wrong. :/

Oh yeah! If 2 of us did the same thing, we might be right! I've asked on Yahoo answers to see, there are proper geniuses there. I'll quote you back when I get the answer. How did the rest of your paper go?

 

[shah zaib]

Oh yeah! If 2 of us did the same thing, we might be right! I've asked on Yahoo answers to see, there are proper geniuses there. I'll quote you back when I get the answer. How did the rest of your paper go?

Lets hope for the best. oh, okay (Y)

Rest of the paper went very well. Other than Q2, i just got the last part of the permutation question wrong.

 

[randomcod]

Lets hope for the best. oh, okay (Y)

Rest of the paper went very well. Other than Q2, i just got the last part of the permutation question wrong.

The 5P something question?! Me too!!!hahaah. I accidentally multiplied by 5 rather than 3, so annoyed - dumb mistake.

 

[Feroz Tahir]

I did the weirdest thing for the Year group question
I did 7C1 x 2C1 x 2C1 (for one of each) x (6C2 + 6C1 x 1C1 + 6C1 x 1C1 + 1C1 x 1C1) (for the other three)
Which is 28 x 28
784

We needed to select 5 people so how can you select 3? Its wrong. Your every possibility is wrong.

 

[samantha liyanage]

Th

that was in q7 I am talking about the one were we had to find the mean and variance

11.7 and the variance was 0.665 something like that

 

[shah zaib]

The 5P something question?! Me too!!!hahaah. I accidentally multiplied by 5 rather than 3, so annoyed - dumb mistake.

Oh.
no, not this one. The one which said that the numbers are less than 1000 and multiples of 5. it was 57, i wrote 49. Didnt consider the possibilities for double and single digit number. :/

 

[Feroz Tahir]

Oh.
no, not this one. The one which said that the numbers are less than 1000 and multiples of 5. it was 57, i wrote 49. Didnt consider the possibilities for double and single digit number. :/

HAHA. And guess what I did? Didn't consider the single digit number xD

 

[randomcod]

We needed to select 5 people so how can you select 3? Its wrong. Your every possibility is wrong.

The first 3 people were chosen at the beginning because they HAD to be there, then you choose the other 2. We don't know who's right, but I'm sticking with my opinion for the moment!

 

[shah zaib]

We needed to select 5 people so how can you select 3? Its wrong. Your every possibility is wrong.

5 people is what he's considered. that's a typo there, he probably meant "for the other 2"

 

[randomcod]

Oh.
no, not this one. The one which said that the numbers are less than 1000 and multiples of 5. it was 57, i wrote 49. Didnt consider the possibilities for double and single digit number. :/

Oh! I did that too in the beginning, then remembered adding 8 due to 5 15 35 45 55 75 85 and 95

 

[randomcod]

5 people is what he's considered. that's a typo there, he probably meant "for the other 2"

What do you mean? Am I right or am I wrong?

 

[shah zaib]

Oh! I did that too in the beginning, then remembered adding 8 due to 5 15 35 45 55 75 85 and 95

i wonder how could i not consider them. I'd probably lose 2 marks here.

 

[Feroz Tahir]

The first 3 people were chosen at the beginning because they HAD to be there, then you choose the other 2. We don't know who's right, but I'm sticking with my opinion for the moment!

Okay. So listen? the question was like There are three groups? One have 7 Members, One have 2 and other have 2. There should be ATLEAST 1 People from each group. We need to select 5.

So firstly to fulfill the requirement of ATLEAST 1 From each we have :

7C3 x 2C1 x 2C1.
Then,
7C1 x 2C2 x 2C2.
Then,
7C2 x 2C1 x 2C2
Then,
7C2 x 2C2 x 2C1.
Adding these all we get the final answer.

 

[randomcod]

Alright guys. I've checked with Yahoo answers. Apparently, the answer is 231. Congrats to all who got it right Also, out of how many marks was this question? Do you think I'll get at least half marks or 1 mark?

 

[shah zaib]

What do you mean? Am I right or am I wrong?

You are right, that's what i think. i just said that you mistyped the "for the other three", it should be "other two".

 

[randomcod]

Okay. So listen? the question was like There are three groups? One have 7 Members, One have 2 and other have 2. There should be ATLEAST 1 People from each group. We need to select 5.

So firstly to fulfill the requirement of ATLEAST 1 From each we have :

7C3 x 2C1 x 2C1.
Then,
7C1 x 2C2 x 2C2.
Then,
7C2 x 2C1 x 2C2
Then,
7C2 x 2C2 x 2C1.
Adding these all we get the final answer.

You're absolutely right

 

[randomcod]

I don't know who's right and who is wrong. It was a 3 / 4 marker so even though 1 of us (probs me) is wrong, we will probably pick 1 / 2 marks. Good luck to all with the rest of the questions! I think I lost 5 marks overall in this paper. Let's hope it doesn't get worse!

 

[Kamihus]

Answers for Paper 62

 

[saadgujjar]

what will be gt