If There are four 2's three 1's and two 0's how many was can you arange them in a 9 Digit number! If There are four 2's, three 1's and two 0's, in how many was can you arrange them in a 9 Digit number! Using Permutations only.
Show your answer is corrrect by counting it in three different ways and getting the same answer each time.
I Got :
p(9,4)*p(5,3)*P(2,2) =9! 
p(9,2)*p(7,3)*P(4,4) =9! 
p(9,3)*p(6,4)*P(2,2) =9!
Am I correct?
 A: $\require{cancel}$
This is a standard problem for using the multinomial coefficient:
$$\binom{9}{4, 3, 2} = \binom{9}{4}\cdot \binom 53\cdot \binom 22 = \dfrac{9!}{4!\cancel{5!}} \cdot \frac{\cancel{5!}}{3!2!} \cdot \frac{\cancel{2!}}{\cancel{2!}} = \frac{9!}{4!\,3!\,2!}=1260$$
We have multiple occurrences of some digits, which can permute without changing the final number. So we need to divide by the number of ways each repeated digit can be permuted.
A: 9 numbers can be arranged in $9!$ ways, but since we cannot distinguish between different instances of 2, 1, and 0, arrangements which differ only by a permutations among these different instances of 2, 1 and 0, should be considered the same. This means that the number of arrangements times the number of ways in which the 2, 1 and 0 can be permuted among each other is the total number of permutations of 9 numbers, ie. 9!. If $N$ is your answer, then we must have $N \times 4! \times 3! \times 2 = 9!$, and so
$$
N = \frac{9!}{4! \times 3! \times 2!} = 1260
$$
