# Amount of variants of a card with 12 fields

I got an additional task from my math teacher today: Theres a card with 12 fields, but only 6 of them are filled with a symbol. In every card, there are 3 star symbols, 2 moon symbols and 1 sun symbol:

The task is to calculate how many different cards of this kind can exist/maximal amount of different cards.
My Solution was $12! - 6!$, because I thought that if I would take one star, I would have 12 places to place and if I would take another one, I would have 11 places to place.
But I already found out that this isnt the right solution, because the places of one of the symbols are interchangeable without making the card look different.
So can someone give me a hint to the right solution?

Any help will be greatly appreciated.

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## 1 Answer

Hint 1: First, imagine all the symbols are distinct. How many ways can you place them? Among these ways, you now consider all the stars to be the same. That means you have counted each arrangement _ times, so divide by that. Do the same for the moons and you are done.

Hint 2: You can go the other direction: How many ways are there to choose which six spaces will be filled? How many ways to pick three of those for stars? How many to pick the places for the moons? Here you multiply.

You should get the same answer either way.

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Thank you, followed Hint 2 and I calculated: (12!)/(6!*6!)*(6!)/(3!*3!)*(3!)/(2!)*1=55440 (first fraction for the 6 filled spaces, 2nd fraction for the stars, 3rd fraction for the moons and the last 1 for the sun). Is it correct? – Tearsdontfalls Feb 28 '13 at 15:45
@Tearsdontfalls: It is hard to look at that expression and figure out what is numerator and what denominator, but it looks right. – Ross Millikan Feb 28 '13 at 16:26