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## How many numbers between 0 and 1 million can be formed using 0,7,8

$1$ million = $1000000$

so digits should be in the range of $1$ to $6$

for filling one digit we have $3$ options (either $0, 7, 8$)

so total possibility for $6$ digits $= 3^6 = 729$

out of which its count $000000$ which is equal to $0,$ can’t be considered for this question.

so, total numbers $= 729 \ – 1 = 728$

Note: I am assuming numbers between $0$ and $1$ million in numbers greater than $0$ and less than $1$ million

@Yaman Sahu I realised this after publishing my comment. But could not delete because there are no such options.

OOh
I think 729 is fine, it is not safe to assume the non-inclusiveness.

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....

If we include 0 we have

$3 + 2*3 + 2*3^{2} + 2*3^{3} + 2*3^{4} + 2*3^{5}$

= 729

Excluding 0 we get 729 -1 = 728

We can also think of this as 6 digit positions each having 3 choices, thus total number of possibilities = $3^{6}$ = 729

Excluding 0 we get $3^{6} - 1$ = 728
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