Lining up pennies, nickels, and dimes

Question

You place 7 coins in a row: 3 identical pennies, 2 identical nickels, and 2 identical dimes. Two rows are the same if the coin type at each position matches. How many distinct arrangements are possible?

Accepted Answer

Coins of the same type are interchangeable, so this is a multiset arrangement. If all 7 coins were distinct there would be 7! orderings, but we overcount by 3! for the pennies, 2! for the nickels, and 2! for the dimes: \frac{7!}{3!\,2!\,2!} = \frac{5040}{6 	imes 2 	imes 2} = \frac{5040}{24} = 210. Takeaway. The same divide-by-repeat-counts rule handles three groups at once: \dfrac{7!}{3!\,2!\,2!} = 210.

Lining up pennies, nickels, and dimes

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