Question

How many different $9$ digit numbers can be made with $3$ fours, $4$ twos and $2$ threes?

Answer 1

According to the question,

Total given numbers = 9 (Three - four, four - twos and 2 - three)

No of ways = $\frac{9!}{(3!\ast 4!\ast 2!)}$ ( Denominator is for repetation of numbers )

$=\frac{(9\ast 8\ast 7\ast 6\ast 5\ast 4!)}{(6\ast 4!\ast 2)}$

$=(9\ast 4\ast 7\ast 5)$

$=1260$