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Mathematics: Post your doubts here!

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(iv) This part is a bit tricky. Here you have to consider a number of possible scenarios.
You are to select exactly 1 N.

So the selections can either contain:
1. Zero As
2. 1 A
3. 2As
OR
4. 3As

N***
^ You are to fill these places shown by asterisks.

First Consider the situation there are no A's. [N***] The possible number of selections from the letters remaining excluding all As will be 3C3.

Then consider the next situation: 1 A. [NA**] The number of possible selections will be : 3C2

Situation for exactly 2 As : [NAA*] This can be done in 3C1 ways.

Situation for exactly 3 As: [NAAA] There is only 1 way to do it, so possible selections are only 1.


Now sum up all these results to get the final answer: 3C3 + 3C2 + 3C1 + 1 = 8

Hence total number of possible selections are 8.
Why can't we write 6C3? Like 6 letters are left other than N so we just calculate possible combinations of the six by 6C3?
 
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Why can't we write 6C3? Like 6 letters are left other than N so we just calculate possible combinations of the six by 6C3?
Because it includes three As, so there'll be repetition. Like let's name these three As as : A1, A2, and A3.
Consider the 2nd and 3rd criteria i mentioned, now see there'll be repition:

N A1 * *
N A2 * *
N A3 * *

^ See, these are actually the same combinations. So it would be wrong!

Get it?
 
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I know it's irrelevant here but please can someone direct me to someone who has given or would b appearing for socio this oct/nov? :(
 
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