How many ways to write the 6 numbers? There are 9 places where each can be written by a number. Numbers 1, 1, 1, 2, 2, and 2 will be written into the place, with conditions: all numbers must be used and each number can only be written once. How many ways to write the 6 numbers?
Many ways are C(9;3).
But we know that there will be double numbers. Is there someone who can help me?
.
 A: The problem can be boiled down into basically two steps.


*

*How many ways can we choose 6 boxes from the 9 to put the numbers in?

*And how many unique ways can we order 1,1,1,2,2,2?


The final answer will just be the product of the answers to these two questions since for every choice of 6 boxes we can put every possible unique combination of the digits in. (Or for every possible unique combination of digits we can put them in any choice of 6 boxes.)
To answer the first one is pretty straightforward, its just $9 \choose 6$ since we are picking 6 out of the 9 boxes to use.
To answer the second question first pretend that instead of 1,1,1,2,2,2 we have 1a,1b,1c,2a,2b,2c. Now to order this latter set of numbers there are 6! ways of doing it because every element is distinct. However, this overcounts since for example 1a,1b,1c,2a,2b,2c and 1b,1c,1a,2c,2b,2a are both counted.
To correct this we need to divide by all the ways we could order the 1's and by all the ways we could order the 2's. There are 3! permutations for three elements which means that there are 3! ways of ordering the 1's and 3! ways of ordering the 2's. Dividing these out we get $\frac{6!}{(3!)(3!)}$.
Now to put them together we just multiply and get ${9 \choose 6} \frac{6!}{(3!)(3!)} = 84(20) = 1680.$
A: Just to nitpick the original problem, I see only two numbers: $1$ and $2.$ It seems you are meant to write each one three times. Moreover, I suppose you are meant to leave three of the places blank.
So you are asked to make an arrangement in the nine places such that three places are filled with $1$, three are filled with $2$, and three are left blank.
Now if the problem were (for example) merely to fill three places with $1$ and leave the other six blank, you would use the binomial coefficient $\binom 93$ or the
binomial coefficient $\binom 96$, depending on whether you are thinking more about the three filled places or the six blank places.
Binomial coefficients are good if there are only choices for each place,
such as put $1$ in it or leave it blank. When there are three or more choices, you can use multinomial coefficients.
Since you want three of each of the three kinds of "filling" ($1,$ $2,$ or blank) in your nine places, the multinomial coefficient that counts the number of ways to fill the places is
$$ \binom{9}{3,3,3} = \frac{9!}{3!3!3!}.$$
Note that you can derive the multinomial coefficient formula by seeing that you have $\binom 93$ ways to choose which places are filled with $1,$ and then for each of those ways to fill places with $1$ you have $\binom 63$ ways to choose what is written in the remaining places. So
$$ \binom{9}{3,3,3} = \binom 93 \binom 63.$$
