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A set "a" has 6 elements.In how many ways can the elements be divided into 2 groups?

Options given: 1) $62$ 2) $31$ 3) $52$ 4) $32$

My Approach:

I divided the items into 2 unequal groups of 4 and 2,5 and 1,6 and 0,3 and 3.

I get

$6!$/$2!$ . $4$!=$15$

$6!$/$3!$ . $3$!=$20$

$6!$/$1!$ . $5$!=$6$

$6!$/$0!$ . $6$!=$1$

total=$42$

Can anyone can give me hint how to solve this problem?

justin takro
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  • $6!/(2!\cdot 4!)$ is not the number of ways to break the elements into two groups. It is the way to break 6 elements into a group of 2 elements and a group of 4 elements. You must also add the ways to break the elements into a group of 1 element and a group of 5 elements as well as a group of 0 elements and a group of 6 elements and breaking the set into two groups of 3 elements. – Michael Burr Sep 27 '15 at 16:15
  • @MichaelBurr By adding all i am getting wrong ans see the edit post please. – justin takro Sep 27 '15 at 17:18
  • @anubhav How is that possible? – justin takro Sep 27 '15 at 17:25

1 Answers1

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Unless there are further restrictions, you have 6 slots. Each slot takes either the value 1 or 2. Therefore, the total number of partitions is 32.

NmdMystery
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Alex
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