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$2$ people are choosing from a menu of 9 food items and must choose $3$ items on the menu each. How many ways are there in which person $1$ and person $2$ choose precisely one of the same menu items?

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Hint: first calculate how many ways person 1 can choose three dishes. Then calculate how many ways person 2 can choose three dishes, with exactly one of them being one of the three that person 1 picked. Also, if this is homework, please mark it as such. – Jonathan Christensen Jan 11 '13 at 22:58
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HINT: Person $1$ can make any of $\binom93$ choices. In order to have exactly one item in common with Person $1$, Person $2$ must choose one of the $3$ items that Person $1$ chose and $2$ of the $6$ items that Person $1$ did not choose. There are $\binom31\binom62$ ways to do that. How should you put these pieces together to get the final result?

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Subtract the two? – fosho Jan 11 '13 at 23:03
@fosho: No, this is a sequence of choices: first Person $1$ makes his choice of $3$ items, and then Person $2$ makes a choice. Look at a simpler problem. Suppose that each was selecting one item, and you wanted the number of ways in which they could select different items. Person $1$ could make his selection in $9$ ways, and then Person $2$ could make his in $8$ ways, for a total of how many different combinations? – Brian M. Scott Jan 11 '13 at 23:06
You add them? for 17? – fosho Jan 11 '13 at 23:07
@fosho: Are there really only $17$ different combinations? Person $1$ can choose $9$ different items. Say he chooses Item $1$; then Person $2$ can choose any of Items $2$-$9$, for a total of $8$ combinations. If Person $1$ chooses Item $2$ instead, Person $2$ can choose Item $1$ or any of Items $3$-$9$, for a total of another $8$ combinations. Can you see that no matter what item Person $1$ chooses, Person $2$ will have $8$ possible choices, and you’ll get another $8$ combinations, for a grand total of $9\cdot 8$ combinations? That’s a lot more than $17$. – Brian M. Scott Jan 11 '13 at 23:11
Ah thanks alot! – fosho Jan 11 '13 at 23:12

There are $\binom{9}{1}$ choices for the common item. For every such choice, there are $\binom{8}{2}$ ways for A to choose the rest of her menu. And for every such choice by A, there are $\binom{6}{2}$ ways for B to choose $2$ items that A did not choose.

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