Answer:
2520
Step-by-step explanation:
If a nine digit number contains exactly 4 one's, 3 two's and 2 three's
We can choose the positions of the 1's in P(9/4) ways.
We can choose the positions of the 2's from the five remaining positions in P(5/3) ways
We can choose the positions of the 3's from the two remaining positions in P(2/2) ways
In this problem order doesnt matter..
= (9!/5!×4!) × (5!/3!×2!) × (2!/2!×0!)
= 2520
