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Find the number of three element subsets $A$ of $\{1,2,...,10\}$ such that $A$ contains at least one even and one odd integer.

I think in this way: First we distribute one even and one odd number to 3 sets.Then there are 4 digits left of which 2 are even and 2 are odd.But how to distribute this 4 digits into 3 sets?

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    $\begingroup$ I don't know why you are distributing numbers to three sets. The question is about three-element sets, and how many different ones there are. $\endgroup$ – Gerry Myerson Apr 11 '13 at 13:30
  • $\begingroup$ I am getting 3 three-element sets and one digit extra.So my question is how to get the three-element set?Is there any other way? $\endgroup$ – Germain Apr 11 '13 at 13:34
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    $\begingroup$ @analysis89, I think you are misunderstanding the problem. Suppose the question didn't have the "such that" part. Then the answer would be the number of ways of choosing 3 things from 10, that is, the binomial coefficient 10-choose-3. We're not partitioning the set into 3-element subsets; we're counting the total number of 3-element subsets that exist. $\endgroup$ – Gerry Myerson Apr 11 '13 at 13:45
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    $\begingroup$ The set $\{1,2,2\}$ does exist. It is a two element set. In particular, $$\{1,2,2\}=\{1,2\}=\{2,1\}$$ $\endgroup$ – Hurkyl Apr 13 '13 at 19:20
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    $\begingroup$ Oh! I was thinking totally blindly.Thanks Hurkyl, Gerry Myerson, Dolma,Douglas S. Stones. $\endgroup$ – Germain Apr 14 '13 at 10:54
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I think you might be able to get what you want by determining how many subsets of three elements you can get (with no constraints), and remove all subsets which only have even or odd integers in it.

Which would be: $$\binom{10}{3}-2\binom{5}{3}=100$$

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This seems to be where you're heading:

  • Step 0: We start with an empty set.

  • Step 1: We add in an odd number; which can be done in $5$ ways.

  • Step 2: We add in an even number; which can be done in $5$ ways regardless of the choice in Step 1.

  • Step 3: We add in any other number; which can be done in $8$ ways regardless of the choices in Steps 1 and 2.

This process generates $5 \times 5 \times 8=200$ (not necessarily distinct) sets. In fact, each set is generated twice. E.g. $\{1,2,3\}$ is generated once when $1$ is picked in Step $1$, and once when $3$ is picked in Step 1. So, we divide the result by $2$.

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2 odd 1 even $\to \binom{5}{2}\binom{5}{1} = 50$ ways

2 even 1 odd $\to \binom{5}{2}\binom{5}{1} = 50$ ways

total $100$

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