Question

How many numbers between 0 and one million can be formed using 0, 7 and 8?

Answer 1

1 million = 1000000.

So, we have to consider up to 6 digit numbers.

For generalization, we consider all less than 6 digit numbers as six-digit numbers with ‘0’ as their beginning digit(s).

All six digits same: (3C1)*(6C6) = 3

Five digits same: (3C1)*(2C1)*[(6!) / {(5!)*(1!)}] = 36

Four digits same and remaining two digits using the others apiece:

(3C1)*(2C2)*[(6!) / {(4!)*(1!)*(1!)}] = 90

Four digits same and remaining two digits using the anyone from the remaining two twice: (3C1)*(2C1)*[(6!) / {(4!)*(2!)}] = 180

Three digits same and remaining three digits using anyone from the remaining two twice: (3C1)*(2C1)*(1C1)*[(6!) / {(3!)*(2!)*(1!)}] = 360

Using any two digits thrice each = (3C2)*[(6!) / {(3!)*(3!)}] = 60.

So, total ‘generalized’ 6-digit numbers can be formed using 0, 7 and 8

= (3+36+90+180+360+60) = 729.

But, these group of numbers include ‘000000’, which being equal to 0, can’t be considered for this question.

So, total numbers can be formed greater than 0 and less than a million using 0, 7 and 8 = 729 - 1 = 728.

Also, there is an easier method for this problem.

This time too, we’ve to consider ‘generalized’ 6-digit numbers. Here for filling any particular digit’s place, we have 3 options.

So, total ‘generalized’ 6-digit numbers can be formed using 0, 7 and 8

= (3^6) = 729.

But, these group of ‘generalized’ 6-digit numbers include ‘000000’, which being equal to 0, can’t be considered for this question.

So, total numbers can be formed greater than 0 and less than a million using 0, 7 and 8 = 729 - 1 = 728.