Question

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

Answer 1

$\mathtt{\huge{ \fbox{ \: \: SOLUTION \: \: }}}$

There are total 7 numbers , of which 2 occur three times and 4 occur two times

The total number of permutations of given digits is

$= \frac{7!}{3! \times 2! } \\ \\ = \frac{7 \times 6 \times 5 \times \cancel4 \times \cancel{3!}}{ \cancel{3! }\times \cancel2 \times 1} \\ \\ = 7 \times 6 \times 5 \times 2 \\ \\ = 420$

Now , these permutations also include the cases when 0 is at extreme left position which makes the number a 6 digit one

Thus , the number of permutations (excluding 0) is

$\sf = \frac{6!}{3! \times 2! } \\ \\ \sf = \frac{ 6 \times 5 \times \cancel4 \times \cancel{3!}}{ \cancel{3! }\times \cancel2 \times 1} \\ \\ \sf = 6 \times 5 \times 2 \\ \\\sf = 60$

Now , The required number of arrangements is the difference between the number of permutations (including 0) and the number of permutations (excluding 0)

So ,

$\sf \hookrightarrow The \: required \: number \: of \: arrangements = 420 - 60</p><p> \\ \\ \sf \hookrightarrow The \: required \: number \: of \: arrangements = 360$

Hence , 360 numbers can be formed by using 1 , 2 , 0 , 2 , 4 , 2 , 4 which is greater than 1000000