# More Fundamental Counting

• Jan 11th 2010, 12:04 PM
kturf
More Fundamental Counting
How many seven-letter words contain at least one X?

How many seven-letter words contain at least two X's?
• Jan 11th 2010, 01:53 PM
Drexel28
Quote:

Originally Posted by kturf
How many seven-letter words contain at least one X?

How many seven-letter words contain at least two X's?

Once again, what. do. you. think?
• Jan 12th 2010, 05:16 AM
kturf
Here's what I think...
At least one X:
26!/6!(26-6)! + 25!/(25-5)! + .... 22!/(22-2)! + 1

At least 2 X's would begin at 25...

Is that right? Or is it something to do with 26^7 minus something?
• Jan 12th 2010, 06:23 AM
Plato
Quote:

Originally Posted by kturf
At least one X:
is it something to do with 26^7 minus something?

Hint: There are $\displaystyle 25^7$ that contain no X.
• Jan 12th 2010, 06:25 AM
kturf
So would it be 25^6 that contain 1 X and 25^5 that contain 2 X's?
• Jan 12th 2010, 06:39 AM
Plato
Quote:

Originally Posted by kturf
So would it be 25^6 that contain 1 X and 25^5 that contain 2 X's?

Absolutely not.
Contains exactly one X: $\displaystyle 7\cdot 25^6$.

In order to receive further help, you must reply with an explanation of why that is the correct answer.
• Jan 12th 2010, 07:43 AM
kturf
For a word with 7 letters and exactly one X, the answer is 7 times 25^6 because you have seven spots available and 25 other possible letters for the 6 spots that are not the X.

So, for one or more X's, the answer would be

7 times 25^6 + 7 times 25^5 + 7 times 25^4 + 7 times 25^3 + 7 times 25^2 + 7 times 25 + 1?
• Jan 12th 2010, 08:07 AM
Plato
Quote:

Originally Posted by kturf
How many seven-letter words contain at least two X's?

$\displaystyle 26^7-25^7-7\cdot 25^6$ WHY?
• Jan 12th 2010, 08:44 AM
kturf
If you could explain why, that would be very helpful.
• Jan 12th 2010, 09:27 AM
Plato
Quote:

Originally Posted by kturf
If you could explain why, that would be very helpful.

$\displaystyle 26^7$ the total.
$\displaystyle 25^7$ contains no X.
$\displaystyle 7\cdot 25^6$ contains exactly one X.
$\displaystyle 26^7-25^7-7\cdot 25^6$ remove those two from the total.
What do you have left?