Numbers less than a million containing the digit 2.

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April 27th 2009, 10:57 AM
Mohron

Numbers less than a million containing the digit 2.

The question is, how many numbers less than 1,000,000 contain the digit 2?

I am getting 600,000. It's too much to type out but I'm using this format.

1 10 10 10 10 10 = 100,000
10 1 10 10 10 10 = ""
10 10 1 10 10 10 = ""
10 10 10 1 10 10 = ""
10 10 10 10 1 10 = ""
10 10 10 10 10 1 = ""
= 600,000
April 27th 2009, 11:44 AM
Plato

Quote:

Originally Posted by Mohron

The question is, how many numbers less than 1,000,000 contain the digit 2?

There are $9^6 - 1$ positive integers less than 1,000,000 that do not contain the digit 2.
So how many do?
April 27th 2009, 12:07 PM
Mohron

Quote:

Originally Posted by Plato

There are $9^6 - 1$ positive integers less than 1,000,000 that do not contain the digit 2.
So how many do?

OK, so 468,560? How do you know there are (9^6) - 1 positive integers that don't contain 2? Like any chance you have work to show?
April 27th 2009, 12:14 PM
Plato

Quote:

Originally Posted by Mohron

OK, so 468,560? How do you know there are (9^6) - 1 positive integers that don't contain 2? Like any chance you have work to show?

Actually it is 468,559. 000000 is not a positive integer.
There are six places that can be filled with nine non-2 digits.
But we don't count 000000.
April 27th 2009, 12:21 PM
Soroban

Hello, Mohron!

Sorry, your approach has a lot of duplication . . .

Quote:

How many numbers less than 1,000,000 contain the digit 2?

I am including 6-digit numbers with leading zeros.
After all, the number $003279$ can represent the number $3279.$

I solved it two ways:
. . [1] counting the numbers containg a 2
. . [2] counting the numbers without a 2 and subtracting from 999,999.
And my answers agreed!

I'll do it head-on (Method 1) . . .

Numbers with exactly one 2:
There are 6 positions for the 2.
The other five digits have 9 choices each: . $9^5$ ways.
. . There are: . $6\cdot9^5 \:=\:{\color{blue}354,\!294}$ numbers with one 2.

Numbers with exactly two 2's:
There are ${6\choose2} = 15$ positions for the two 2's.
The other four digits have 9 choices each: . $9^4$ ways.
. . There are: . $15\cdot9^4 \:=\:{\color{blue}98,\!415}$ numbers with two 2's.

Numbers with exactly three 2's:
There are ${6\choose3} = 20$ positions for the three 2's.
The other three digits have 9 choices each: . $9^3$ ways.
. . There are: . $20\cdot9^3 \:=\:{\color{blue}14,\!580}$ numbers with three 2's.

Numbers with exactly four 2's:
There are ${6\choose4} = 15$ positions for the four 2's.
The other two digits have 9 choices each: . $9^2$ ways.
. . There are: . $15\cdot9^2 \:=\:{\color{blue}1,\!215}$ numbers with four 2's.

Numbers with exactly five 2's:
There are ${6\choose5} = 6$ positions for the five 2's.
The other digit has 9 choices: . $9$ ways.
. . There are: . $6\cdot9 \:=\:{\color{blue}54}$ numbers with five 2's.

Numbers with exactly six 2's:
There is one number with six 2's (namely, 222,222).

Therefore, the number of 6-digit numbers that contain a 2 is:

. . $354,\!294 + 98,\!1415 + 14,\!580 + 1,\!215 + 59 + 1 \;=\;\boxed{468,\!559}$
April 27th 2009, 12:22 PM
Mohron

Quote:

Originally Posted by Plato

Actually it is 468,559. 000000 is not a positive integer.
There are six places that can be filled with nine non-2 digits.
But we don't count 000000.

(9^6) = 531,441
(9^6) - 1 = 531,440

1,000,000 - 531,440 = 468,560

So it is not 468,559 correct? I took into account the (- 1)

Thanks for the help by the way. Greatly appreciated!
April 27th 2009, 12:31 PM
Plato

Quote:

Originally Posted by Mohron

(9^6) = 531,441
1,000,000 - 531,440 = 468,560

There are only 999,999 positive integers less than 1,000,000.