Question

How many numbers greater than $1000000$ can be formed by using the digits $1,2,0,2,4,2,4$

Answer 1

Case 1- If number begin with 1

The remaining digit to be arranged will be $0,2,2,2,4,4$

Here three $2s$ and two $4s$

Since digit are repeating hence we use this formula$=\frac{n!}{p_1!p_2!p_3!}$

Number of remaining digit$=6$

Thus, $n=6$

Since three $2s$ and two $4s$

and $p_1=3,p_2=2$

The number of numbers beginning with 1$=\frac{n!}{p_1!p_2!}$

$=\frac{6!}{3!2!}$

$=\frac{6\times 5\times 4\times 3!}{3!\times 2\times 1}=60$

Case 2- If number begin with 2

The remaining digit to be arranged will be $1,0,2,2,4,4$

Here two $2s$ and two $4s$

Since digit are repeating hence we use this formula$=\frac{n!}{p_1!p_2!p_3!}$

Number of remaining digit$=6$

Thus, $n=6$

Since two $2s$ and two $4s$

and $p_1=2,p_2=2$

The number of numbers beginning with 2$=\frac{n!}{p_1!p_2!}$

$=\frac{6!}{2!2!}$

$=\frac{6\times 5\times 4\times 3\times 2\times 1}{(2\times 1)\times (2\times 1)}=180$

Case 3- If number begin with 4

The remaining digit to be arranged will be $1,0,2,2,2,4$

Here three $2s$

Since digit are repeating hence we use this formula$=\frac{n!}{p_1!p_2!p_3!}$

Number of remaining digit$=6$

Thus, $n=6$

Since three $2s$

and $p_1=3$

The number of numbers beginning with 4$=\frac{n!}{p_1!}$

$=\frac{6!}{3!}$

$=\frac{6\times 5\times 4\times 3!}{3!}=120$

Required number$=$ Number starting with 1 + Number starting with 2 + Number starting with 4

$=60+180+120$

$=360$