Respuesta :

kritter1
Let's look at the problem; 

There are 7 digits in 3,000,000 and 7 given digits. However, the numbers starting with either 6 or 4 will be greater than 3,000,000. 

If the first digit is 4, 6!/(2!3!) = 60 <--- I knew that there would be 3 repetitions of 6 and 2 repetitions of 2, which I divided away. 

If the first digit is 6, 6!/(2!2!)= 180 <-- repeat as above 

The first digit can be 6 or 4, therefore, we add the possible ways. 

60+180=240 possible permutations 

Hope I helped :) 

There are 240 numbers greater than 3,000,000 can be formed from permutations of 1, 2, 2, 4, 6, 6, 6

The number to be formed that is greater than 3000000 has 7 digits

Also the given digits in the problem are 1,2,2,4,6,6,6 .

The number of digits given in the problem are also 7.

If we have to form the number that is greater than 3000000 it must start either by 4 or 6 .

Case 1 : When first digit is 4 .

Remaining there are 6 places which can be filled in 6 ! ways.

Also there are  two repetitions of digit  2  and three repetitions of digit 6  

So the number of ways possible to write the number in this  case will be

[tex]\rm 6! /2! 3! = 60[/tex]

Case 2  : When first digit is 6 .

Remaining there are 6 places which can be filled in 6 ! ways.

Also there are  two repetitions of digit 2  and two repetitions of digit 6.

So the number of ways possible to write a number in this case will be

[tex]\rm 6! /2! 2! = 180[/tex]

So the total  number of ways possible to write a number that is greater than 3000000 is  180 + 60 = 240

So there are 240 numbers greater than 3,000,000 can be formed from permutations of 1, 2, 2, 4, 6, 6, 6

For more information please refer to the link given below

https://brainly.com/question/23283166

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