How many even 8-digit numbers less than 40 000 000 can be formed using the digits 1,

How many even 8-digit numbers less than 40 000 000 can be formed using the digits 1, 2, 3, 3, 3, 6, 6, 8? Show your work and explain your reasoning.


Should I have only included 3 once as a first number since it is a duplicate?

JSmith said:

How many even 8-digit numbers less than 40 000 000 can be formed using the digits 1, 2, 3, 3, 3, 6, 6, 8?
Show your work and explain your reasoning.

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In order to have the number less than 40 000 000,
the digit at the beginning of the number must be less than > > > or equal to 4. < < <

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No, the beginning digit must be strictly less than 4.

JSmith said:

How many even 8-digit numbers less than 40 000 000 can be formed using the digits 1, 2, 3, 3, 3, 6, 6, 8? Show your work and explain your reasoning.

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This is just a matter of casing it out.
The unit digit can be \(\displaystyle 2,~6,~8\). Things are complicated by two sixes.

The first digit can be \(\displaystyle 1~,2,~3\). Things are complicated by three threes.

A number of the form \(\displaystyle 3XXXXXX6\) can be made in \(\displaystyle \frac{6!}{2}\) ways.

A number of the form \(\displaystyle 1XXXXXX8\) can be made in \(\displaystyle \frac{6!}{2!\cdot 3!}\) ways.

Now you have to consider all the impossibles.

Hello, JSmith!

I haven't checked your work,
. . but I came up with the same answer.
I solved it in an exhaustive way, checking all eight cases.

How many even 8-digit numbers less than 40 000 000 can be formed
using the digits 1, 2, 3, 3, 3, 6, 6, 8? .Show your work and explain your reasoning.

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Begins with 1, ends in 2: .\(\displaystyle 1\;\_\;\_\;\_\;\_\;\_\;\_\;2\)
. . The middle digits \(\displaystyle \{3,3,3,6,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{3!2!} = 60\) ways.

Begins with 1, ends in 6: ..\(\displaystyle 1\;\_\;\_\;\_\;\_\;\_\;\_\;6\)
. . The middle digits \(\displaystyle \{2,3,3,3,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{3!} = 120\) ways.

Begins with 1, ends in 8: ..\(\displaystyle 1\;\_\;\_\;\_\;\_\;\_\;\_\;8\)
. . The middle digits \(\displaystyle \{2,3,3,3,6,6\}\) can be arranged in \(\displaystyle \frac{6!}{3!2!} = 60\) ways.

Begins with 2, ends in 6: ..\(\displaystyle 2\;\_\;\_\;\_\;\_\;\_\;\_\;6\)
. . The middle digits \(\displaystyle \{1,3,3,3,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{3!} = 120\) ways.

Begins with 2, ends in 8: ..\(\displaystyle 2\;\_\;\_\;\_\;\_\;\_\;\_\;8\)
. . The middle digits \(\displaystyle \{1,3,3,3,6,6\}\) can be arranged in \(\displaystyle \frac{6!}{3!2!} = 60\) ways.

Begins with 3, ends in 2: ..\(\displaystyle 3\;\_\;\_\;\_\;\_\;\_\;\_\;2\)
. . The middle digits \(\displaystyle \{1,3,3,6,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{2!2!} = 180\) ways.

Begins with 3, ends in 6: ..\(\displaystyle 3\;\_\;\_\;\_\;\_\;\_\;\_\;6\)
. . The middle digits \(\displaystyle \{1,2,3,3,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{2!} = 360\) ways.

Begins with 3, ends in 8: ..\(\displaystyle 3\;\_\;\_\;\_\;\_\;\_\;\_\;8\)
. . The middle digits \(\displaystyle \{1,2,3,3,6,6\}\) can be arranged in \(\displaystyle \frac{6!}{2!2!} = 180\) ways.


Therefore, there are:
. . \(\displaystyle 60+120+60+120+60+180+360+180 \:=\:1140\) such numbers.

Thank you very much... This confirms that I believe I have the correct answer. Take care!

soroban said:

Begins with 1, ends in 2: .\(\displaystyle 1\;\_\;\_\;\_\;\_\;\_\;\_\;2\)
. . The middle digits \(\displaystyle \{3,3,3,6,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{3!2!} = 60\) ways.

Begins with 1, ends in 6: ..\(\displaystyle 1\;\_\;\_\;\_\;\_\;\_\;\_\;6\)
. . The middle digits \(\displaystyle \{2,3,3,3,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{3!} = 120\) ways.

Begins with 1, ends in 8: ..\(\displaystyle 1\;\_\;\_\;\_\;\_\;\_\;\_\;8\)
. . The middle digits \(\displaystyle \{2,3,3,3,6,6\}\) can be arranged in \(\displaystyle \frac{6!}{3!2!} = 60\) ways.

Begins with 2, ends in 6: ..\(\displaystyle 2\;\_\;\_\;\_\;\_\;\_\;\_\;6\)
. . The middle digits \(\displaystyle \{1,3,3,3,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{3!} = 120\) ways.

Begins with 2, ends in 8: ..\(\displaystyle 2\;\_\;\_\;\_\;\_\;\_\;\_\;8\)
. . The middle digits \(\displaystyle \{1,3,3,3,6,6\}\) can be arranged in \(\displaystyle \frac{6!}{3!2!} = 60\) ways.

Begins with 3, ends in 2: ..\(\displaystyle 3\;\_\;\_\;\_\;\_\;\_\;\_\;2\)
. . The middle digits \(\displaystyle \{1,3,3,6,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{2!2!} = 180\) ways.

Begins with 3, ends in 6: ..\(\displaystyle 3\;\_\;\_\;\_\;\_\;\_\;\_\;6\)
. . The middle digits \(\displaystyle \{1,2,3,3,6,8\}\) can be arranged in \(\displaystyle \frac{6!}{2!} = 360\) ways.

Begins with 3, ends in 8: ..\(\displaystyle 3\;\_\;\_\;\_\;\_\;\_\;\_\;8\)
. . The middle digits \(\displaystyle \{1,2,3,3,6,6\}\) can be arranged in \(\displaystyle \frac{6!}{2!2!} = 180\) ways.


Therefore, there are:
. . \(\displaystyle 60+120+60+120+60+180+360+180 \:=\:1140\) such numbers.

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