Why does ways to pick $12$ elements from $5$ different types not equal to $5^{12}$?

The question is how many ways you can pick $$12$$ donuts from $$5$$ different varieties. I thought I can just pick $$1$$ donut at a time. Each time I have $$5$$ choices for $$1$$ donut so it would be $$5^{12}$$ ways but it’s wrong.

• I assume it is because order doesn't matter. Picking Raspberry, Raspberry, Old Fashioned, Glazed, Buttermilk is the same thin as picking Buttermilk, Raspberry, Old Fashioned, Raspberry, Glazed. Commented Feb 19, 2020 at 17:16
• Welcome to Mathematics Stack Exchange. This should be tagged combinatorics not ordinary-differential-equations Commented Feb 19, 2020 at 17:16
• It's kind of frustrating as in combinatorics classes they tend to intuitively expect phrases like "number of ways" to be clear in context. The ways to choose donuts implies that you put the donuts in the bag and order doesn't matter. But the number of ways to select speakers for a conference, order might matter. Ways to put rings on your fingers ... does it matter which rings go above or under each other? Anyway... if you want to line the donuts on a tray one after another. The answer is $5^{12}$. But if you are dumping them in a bag, it isn't. Commented Feb 19, 2020 at 17:26
• @fleablood Indeed, in a literal interpretation "ways to pick donuts" might lead to even more posibilities (picking them with your left vs right hand, for example) :( Commented Feb 19, 2020 at 18:09
• Or may likely "I want that cherry glazed with the nice bubbly bit on the side but not that other cherry glazed with the crust brown bit on top". It's a mixed blessing. On the one hand perhaps more than any other math branch combinatorics is practical real world so it should be intuitive. But there should be a point where we say "now that you understand the difference lets formally and consistantly define the many ways we can 'pick things'". Commented Feb 19, 2020 at 18:28

So the number of ways to display twelve donuts in a row from $$12$$ different types is indeed $$5^{12}$$.