combination?

Question

How many whole numbers less than 1000 contain at least one 2 but no 3?

Answer 1

It has 3 digits
1*9*9 = 81 --- hundreds digit is 2
7*(1*8*2 + 1) = 119 --- hundreds digit is not 2

Expound:
In the 1st case, the rest of digits have 9 choices (remove 3).
In the 2nd case, hundreds digit must not be 0,2 or 3. One of the other digits must be a 2, the other one can be any non-'3'.

Sum: 200 of the 3 digit numbers satisfy.



d:

Edit: Ha ha , I forgot the 2digit and the digit 2.
2digit: 16
digit 2: 1
add them: grand total: 217.
(RamblinWill is correct!)

Answer 2

assuming your looking at integers from 0 to 999, then what we really need to do is count the number of 3 digit "strings" that can be written with the "characters" [0 1 2 4 5 6 7 8 9 ] less the number of 3 digit strings that can written with [0 1 4 5 6 7 8 9].

This will count the answer to your question because each number from 0 to 999 is represented by One 3 digit string (0 = "000") and each 3 digit string is represented by One number from 0 to 999, and because the principle of inclusion/exclusion says the # of a Set with at least One '2' is the same as the difference between the # of the Set and the # set with no '2's.

# of 3 digit strings with [0 1 2 4 5 6 7 8 9] = 9^3
# of 3 digit strings with [0 1 4 5 6 7 8 9] = 8^3
# of 3 digit strings with at least One 2 but no 3's = 9^3 - 8^3

So the final answer is 217

Answer 3

Given that one digit is a 2 and no digit can be a 3 then there will be just 1 single digit number (2).
There will be 16 two-digit numbers.
The three digit numbers will be 7*19+90= 223.
So total is 223+16+1=240

Answer 4

One digit: 1; namely {2}
Two digit: 16; namely {20,21,22,24,25,26,27,28,
29,12,42,52,62,72,82,92}
Three digit with one 2: 3*8*8=192
Three digit with two 2's: 3*8=24
Three digit with three 2's: 1

1+16+192+24+1
=234

Answer 5

234. there are 17 in each set of 100, except for 200 where there are 81 because the thirties don't count and the ones that end in 3 don't count.