# Min Number of Values from {1,2,…,9} Such that diff of 2 picked values is 5

This is a question from Shcaum's whose answer I don't understand. Our textbook has 2 pages on the pigeonhole principle and I'm having quite a bit of difficulty with it.

Give the set ${1,2,...,9}$ find how many members must be chosen to guarantee that at least one pair has a difference of 5.

There are $\binom{9}{2}=36$ possible pairs, of which exactly 4 $\{ \{9,4\},\{8,3\},\{7,2\},\{6,1\} \}$ result in a difference of 5. They then add $\{5\}$ to the set and say there are 5 pigeonholes which requires picking a minimum of 6 numbers.

Could someone explain this?

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I guess Shcaum's is a typo and you mean Schaum's outlines. There's plenty of book in this series, so perhaps you could be more specific, if you want to mention the book you're using. –  Martin Sleziak Jan 8 '12 at 10:57

You can certainly pick $5$ numbers to not have the difference of $5$ ; just choose $\{ 1,2,3,4,5\}$. But you can't choose $6$, since if you consider $\{ \{1,6\}, \{2,7\}, \{3,8\}, \{4,9\}, \{5\} \}$, by the pigeonhole principle if you pick $6$ numbers out of $5$ disjoint subsets there must be one amongst which you've picked $2$ numbers, hence a pair.