# Combinations Help

• Nov 2nd 2008, 03:31 PM
ezwind72
Combinations Help
This is a large word problem. I have stared at it for 2 days and I just can't wrap my mind around where to start. Any help in the right direction would be greatly appreciated!

Suppose that for a given computer, a password must satisfy each of the following conditions:

1) The password must be between 2 and 6 characters
2) The first character must be a letter and the last character must be a digit between 0 & 9.
3) for passwords with 3 or more characters, all characters except for the first or last may be either a letter or a digit.
4) repetition is allowed.
5) they are not case sensitive.

How many passwords are possible for this computer?

I did it like this:

(26C4)+(36C3)+(36C2)+(10C1)=26300. This does not look right to me!!
• Nov 2nd 2008, 07:01 PM
Soroban
Hello, ezwind72!

These are not combinations!
With a password, the order of the characters is important.

Quote:

Suppose that a password must satisfy the following conditions:

1) The password must be between 2 and 6 characters
2) The first character must be a letter and the last character must be a digit (0 - 9).
3) All characters except for the first or last may be either a letter or a digit.
4) Repetition is allowed.
5) They are not case sensitive.

The first must be a letter: 26 choices.
The last must be a digit: 10 choices.
The others must be a letter or a digit: 36 choices.

$\begin{array}{c|c} \text{Length} & \text{Choices} \\ \hline
2 & 26\cdot10 \\ 3 & 26\cdot36\cdot10 \\ 4 & 26\cdot36\cdot36\cdot10
\\ 5 & 26\cdot36\cdot36\cdot36\cdot10 \\
6 & 26\cdot36\cdot36\cdot36\cdot36\cdot10 \end{array}$

Total: . $260 + 260\cdot36 + 260\cdot36^2 + 260\cdot36^3 + 260\cdot36^4$

. . $= \;260\left(1 + 36 + 36^2 + 36^3 + 36^4\right) \;=\;260\,\frac{36^5-1}{36-1}$

. . $=\;260\,(1,\!727,\!605) \;=\;\boxed{449,\!177,\!300}$

• Nov 2nd 2008, 07:13 PM
ezwind72
Quote:

Originally Posted by Soroban
Hello, ezwind72!

These are not combinations!
With a password, the order of the characters is important.

The first must be a letter: 26 choices.
The last must be a digit: 10 choices.
The others must be a letter or a digit: 36 choices.

$\begin{array}{c|c} \text{Length} & \text{Choices} \\ \hline
2 & 26\cdot10 \\ 3 & 26\cdot36\cdot10 \\ 4 & 26\cdot36\cdot36\cdot10
\\ 5 & 26\cdot36\cdot36\cdot36\cdot10 \\
6 & 26\cdot36\cdot36\cdot36\cdot36\cdot10 \end{array}$

Total: . $260 + 260\cdot36 + 260\cdot36^2 + 260\cdot36^3 + 260\cdot36^4$

. . $= \;260\left(1 + 36 + 36^2 + 36^3 + 36^4\right) \;=\;260\,\frac{36^5-1}{36-1}$

. . $=\;260\,(1,\!727,\!605) \;=\;\boxed{449,\!177,\!300}$

I understand where you are coming from with the first being 26 choices and the last being 10. I am confused why it would be 26x10 though. I just loathe these word problems! I appreciate all your help though.
• Nov 2nd 2008, 07:16 PM
ezwind72
BTW, you said it isn't a combination. Does that make it a permutation?
• Nov 2nd 2008, 08:04 PM
Rheanna
ooh hope I get this one as an extra credit ;p (Rofl)