combinations & permutations

C

carebear

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Consider the numbers between 100 and 400. How many of these numbers are even and have no repetition of digits?
Case 1: 2 or 3 in first position and 0 in last position: 2x8x1 = 16
Case 2: 1 or 3 in first position and 2 in last position: 2x8x1 = 16
Case 3: 1,2 or 3 in first position and 4,6,8 in last position: 3x8x3 = 72
Total:104....BUT the answer is 112. What am I missing?
 
Consider the numbers between 100 and 400. How many of these numbers are even and have no repetition of digits?
Case 1: 2 or 3 in first position and 0 in last position: 2x8x1 = 16
Case 2: 1 or 3 in first position and 2 in last position: 2x8x1 = 16
Case 3: 1,2 or 3 in first position and 4,6,8 in last position: 3x8x3 = 72
Total:104....BUT the answer is 112. What am I missing?

102, 104, 106, 108, 120, 124, 126, 128, 130, 132, 134, 136, 138, 140, ...
200, 202, 204, 206, 208, 210, 214, 216, 218, 230, 234, 236, 238, 240,...
200, 302, 304, 306, 308, 310, 312, 314, 316, 318, 320, 324, 326, 328,...
300, 302, 304, ...

Did I misinterpret your question?

I read it as though you are seeking all the "even" numbers between 100 and 400 that have no repeating digits within the number.
 
no you did not misinterpret situation.....that is correct....what am I doing wrong?
 
no you did not misinterpret situation.....that is correct....what am I doing wrong?

I made an error in the original table.

Complete the table of values I started.

102, 104, 106, 108, 120, 124, 126, 128, 130, 132, 134, 136, 138, 140, 142, 146, 148, 150, 152, 154, 156, 158, 160, 162, 164, 168, 170, 172, 174, 176, 178, 180, 182, 184, 186, 190, 192, 194, 196, 198 or 40 numbers.

I'll let you complete the remaining numbers.

200, 202, 204, 206, 208, 210, 214, 216, 218, 230, 234, 236, 238, 240,..............290, 294, 296, 298

300, 302, 304, 306, 308, 310, 312, 314, 316, 318, 320, 324, 326, 328,..............392, 394, 396, 398
 
Thank you, but is there anyone else able to help.....while this will work, I don't want to count this out....it would become VERY time consuming if the question were changed to between 100 and 900, etc. Thanks anyways.

Can anyone else please help?
 
Hello, carebear!

Consider the numbers between 100 and 400.
How many of these numbers are even and have no repetition of digits?
Click to expand...

The first digit is \(\displaystyle \{1,2,3\}.\)
The last digit is: \(\displaystyle \{0,2,4,6,8\}.\)


First digit is 1: .\(\displaystyle 1\;\_\;\_\)
. . The last digit is even: 5 choices.
. . The middle digit is any of the remaining 8 digits.
\(\displaystyle \text{There are: }\:5\cdot 8 \:=\:\boxed{40}\) numbers beginning with 1.

First digit is 2: .\(\displaystyle 2\;\_\;\_\)
. . The last digit is even: 4 choices. .{0, 4, 6, 8}
. . The middle digit is any of the remaining 8 digits.
\(\displaystyle \text{There are: }\:4\cdot8 \:=\:\boxed{32}\) numbers beginning with 2.

First digit is 3: .\(\displaystyle 3\;\_\:\_\)
. . The last digit is even: 5 choices.
. . The middle digit is any of the remaining 8 digits.
\(\displaystyle \text{There are: }\:5\cdot8 \:=\:\boxed{40}\) numbers beginning with 3.

\(\displaystyle \text{Therefore, there are: }\:40 + 32 + 40 \:=\:\boxed{112}\text{ such numbers.}\)

 
Thank you so much for simplifying it....I hate when I "over" think it.....or maybe I should say "under" think......
 
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