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Mathematics: Post your doubts here!

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Spiceangel The Sarcastic Retard farhan141 Awesome12 *Le me* Akshajistari mistique_bee Copy Cat muhammadali233 qwertypoiu
 
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Anyone please try to solve these parts and explain.
ON13 Q6iv V62
MJ12Q5iii V62
ON11Q3b V62
ON11Q2ii V62
MJ11Q4iii V62
MJ09Q4iii V62
 
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Anyone please try to solve these parts and explain.
ON13 Q6iv V62
REMEMBRANCE

Here you are told that the selection does not contain Rs and Ms, so eliminate these letters from the word; you'll be left with:
EEE BNCA

Now the selection can contain either 3Es OR 2 Es. So consider these possibilities separately:
1) Selection contains 3 Es.

*EEE
^the asterisk here can be replaced by any of the other 4 letters, so no. of ways this can be done is 4C1

2) Selection contains 2Es

**EE
^again the asterisks can be replaced by any two letters from B,N,C or A. So no. of possible selections are: 4C2

Add these two results, you'll get the total no. of possible selections, which contain atleast 2 Es: 4C1 + 4C2 = 10 Ans.
 
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MJ12Q5iii V62
There are 8 questions in Section A, and 3 questions in Section B.

First consider the situation, that both Q1 & Q2 are included. Then you'll have to select 4 more questions from the remaining 9 questions. This can be done in 9C4 ways.

Now, consider the situation that neither Q1 is included nor Q2 is included, so you'll have to choose all of the 6 questions from the remaining 9 Questions. This will be done in 9C6 ways.

Sum these two, you'll get the answer: 9C4 + 9C6 = 210 selections.
 
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REMEMBRANCE

Here you are told that the selection does not contain Rs and Ms, so eliminate these letters from the word; you'll be left with:
EEE BNCA

Now the selection can contain either 3Es OR 2 Es. So consider these possibilities separately:
1) Selection contains 3 Es.

*EEE
^the asterisk here can be replaced by any of the other 4 letters, so no. of ways this can be done is 4C1

2) Selection contains 2Es

**EE
^again the asterisks can be replaced by any two letters from B,N,C or A. So no. of possible selections are: 4C2

Add these two results, you'll get the total no. of possible selections, which contain atleast 2 Es: 4C1 + 4C2 = 10 Ans.
Oh right. Thanks.
 
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for the part when both questions are to be chosen from part A, we have 4 options to choose right. Why aren't these the correct options?
4C4*3C0+ 4C3*3C1+ 4C2*3C2+ 4C1*3C3 since questions from part B are also chosen at the same time.
 
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There're two mistakes. Firstly, you aren't considering both of the scenarios i.e. when both questions are included and when neither of them is included. You are missing the scenario when none of those two questions are selected. Secondly, When two first questions are chose, you are left with 6 options to choose from part A, not 4.

So going with your method, the correct solution would have been like this:
(6C4*3C0+ 6C3*3C1+ 6C2*3C2+ 6C1*3C3) + (6C6 * 3C0 + 6C5 * 3C1 + 6C4 * 3C2 + 6C3 * 3C3)
= 210 Ans.

Get it?
 
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There're two mistakes. Firstly, you aren't considering both of the scenarios i.e. when both questions are included and when neither of them is included. You are missing the scenario when none of those two questions are selected. Secondly, When two first questions are chose, you are left with 6 options to choose from part A, not 4.

So going with your method, the correct solution would have been like this:
(6C4*3C0+ 6C3*3C1+ 6C2*3C2+ 6C1*3C3) + (6C6 * 3C0 + 6C5 * 3C1 + 6C4 * 3C2 + 6C3 * 3C3)
= 210 Ans.

Get it?
Yeah I know that was incomplete, I was talking about the first scenario only and I think I misread it or like didn't pay attention and thought we had to choose 6. Thanks anyway.
 
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In part 1 i got the answer to number but the probability is confusing. Also why is variance being treated as p in part 2?image.jpeg
 
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Can someone please try these parts and explain
ON 2011 p33 q 3 ii[I get two more values that are not in the ms]
ON 2011 p33 q6 ii
ON 2011 p33 8 ii and iii
ON 2011 p33 q9i[why do we use scalar product if its parallel don't we take ratios?] and q9 iii
ON 2011 p33 q10:confused::confused::confused:

that's almost the whole paper but can't help being dumb:cry::cry:
 
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In the (ii) part first, separate the consonant and the vowels from the given word. You will get:

VOWELS: A, A, A, I
CONSONANTS: T, Z, N, N

Let VOWELS be represented by '*' and Consonants by 'O'

You are given in the question that the first letter is a Consonant, the second vowel and so on alternately, so arrange them in this way:

O * O * O * O *

^ These are the possible positions for the VOWELS and CONSONANTS. So now arrange them in these places. You can arrange the Consonants in this 4 places in 4! ways. There are two Ns so you'll divide 4! with 2!.
Similarly, the vowels can be arranged in 4! ways. There are 3As, so you'll divide by 3!.

Hence, the overall answer will become:

4!/2! * 4!/3!
 
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