Permutation Word Problem

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• Sep 10th 2009, 12:54 PM
sharkman
Permutation Word Problem
How can I solve this problem without using the fundamental counting principle?

The 7-digit phone numbers in a city all have 661 as the first three digits. How many phone numbers are possible?

I was told this is a permutation NOT a combination word problem.

In that case, can I apply nPn or nPr. If not, why not?
• Sep 10th 2009, 01:42 PM
Matt Westwood
Since you know what the first three numbers are, it boils down to how many possible ways the last 4 digits can be.

You can't apply nCn or nPn because there's no constraint that the digits must be unique.
• Sep 14th 2009, 06:32 AM
sharkman
you said...
Quote:

Originally Posted by Matt Westwood
Since you know what the first three numbers are, it boils down to how many possible ways the last 4 digits can be.

You can't apply nCn or nPn because there's no constraint that the digits must be unique.

You said:

"You can't apply nCn or nPn because there's no constraint that the digits must be unique."

What do you mean by that statement?
• Sep 14th 2009, 07:02 AM
Plato
Quote:

Originally Posted by sharkman
How can I solve this problem without using the fundamental counting principle?The 7-digit phone numbers in a city all have 661 as the first three digits. How many phone numbers are possible?

It is impossible to do this problem without using the fundamental counting principle?.
So using it the answer is simply \$\displaystyle 10^4\$.
• Sep 14th 2009, 10:16 AM
Matt Westwood
Quote:

Originally Posted by sharkman
You said:

"You can't apply nCn or nPn because there's no constraint that the digits must be unique."

What do you mean by that statement?

To use nCn or nPn you must assume that if you have one 3, for example, you can't have any more 3's. Here you can have, for example, 3333 or whatever.