2
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What is the number of subsets at size $k$ of the set $\{1,\ldots,n\}$ such that if a subsets contains $2$ it does not contain $1$?

So think about that. We have $n$ numbers and $k$ subgroups. If $2$ contains so have groups on size $(k-1)n +$, but how to start with that.

Consider the set $A=\{1,\ldots,n\}$ .Count the number of subsets of $A$ of cardinality $k$. How many subsets of cardinality on size $k$ do contain the number $2$ and not $1$?

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  • $\begingroup$ In what sense do you mean 'subgroups' (probably not algebraically?) It would help if you defined more precisely what you are trying to count here. $\endgroup$ – Beckham Myers Aug 18 '19 at 21:52
  • $\begingroup$ Is this a group (a set together with a binary operation that is associative, has a neutral element, and inverses), or do you mean set and subsets (which some people sometimes translate as “group” and “subgroups”)? $\endgroup$ – Arturo Magidin Aug 18 '19 at 22:26
  • $\begingroup$ yes i mean that - set and subsets (which some people sometimes translate as “group” and “subgroups” . i need to count how much sub group i have on size k of the grop {1...n} $\endgroup$ – thebesthere Aug 18 '19 at 22:28
  • $\begingroup$ @thebesthere. Mathematically the translation is a blunder. $\endgroup$ – William Elliot Aug 18 '19 at 22:50
  • $\begingroup$ maybe now its clear .. thanks :) Consider the set A={1....n} .Count the number of subsets of A of cardinality k. How many subsets of cardinality on size k do contain the number 2 and not 1 $\endgroup$ – thebesthere Aug 18 '19 at 23:09
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HINTS:

You need the number of subsets of size $k$ that do not contain both $1$ and $2$. This is the number of subsets of size $k$ minus the number of subsets of size $k$ that contain both $1$ and $2$. If a subset of size $k$ contains both $1$ and $2$, then it must contain $k-2$ elements other than $1$ and $2$.

Can you finish it now?

EDIT

The answer is $${n\choose k}-{n-2\choose k-2}$$

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  • $\begingroup$ SO the answer is ( n−2 k−1 ) $\endgroup$ – thebesthere Aug 19 '19 at 13:24
  • $\begingroup$ @thebesthere Do you mean ${n-2\choose k-1}$? $\endgroup$ – saulspatz Aug 19 '19 at 13:27
  • $\begingroup$ yes but why the answer is ( n k )−( n−2 k−2 ) if u can explain plz =] $\endgroup$ – thebesthere Aug 19 '19 at 14:31
  • $\begingroup$ The first paragraph explains it. Which part do you have trouble understanding? By the way, you can format ${n\choose k}$ as ${n\choose k}$ $\endgroup$ – saulspatz Aug 19 '19 at 14:49
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    $\begingroup$ It is not true that ${n \choose k} - {n-2 \choose k-2} = {n-2 \choose k-1}$. $\endgroup$ – Michael Lugo Aug 19 '19 at 15:04

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