# Thread: A question related to Permuation and Combination

1. ## A question related to Permuation and Combination

(b) A computer password, which must contain 6 characters, is to be chosen from the following
10 characters:

Symbols ? ! *
Numbers 3 5 7
Letters W X Y Z

Each character may be used once only in any password. Find the number of possible passwords
that may be chosen if
(i) there are no restrictions, [1]

(iii) each password must contain at least one symbol. [3]

For no iii) each password must contain at least one symbol, I did 3 x 7 x 6 x 5 x 4 x 3 + 3 x 2 x 7 x 6 x 5 x 4 + 3 x 2 x 1 x 7 x 6 x 5 , is this correct?

2. ## Re: A question related to Permuation and Combination

A question.
I have a question about quartiles. Can I organize from lowest to highest cualitatives variables through a weight assigned to them?
For example
John 4
Alex 3
Mary 2
Bob 1 and so if i have a sample: 3 3 3 3 3 1 1 1 2 2 2 2 2 4 4 , can I find the quartiles yet I find for example Q2= 2,5? Thank you

3. ## Re: A question related to Permuation and Combination

can we get post #2 moved to it's own thread?

4. ## Re: A question related to Permuation and Combination

Originally Posted by Zhen64
(b) A computer password, which must contain 6 characters, is to be chosen from the following
10 characters:
Symbols ? ! *
Numbers 3 5 7
Letters W X Y Z
Each character may be used once only in any password. Find the number of possible passwords
that may be chosen if
(i) there are no restrictions, [1]
(iii) each password must contain at least one symbol. [3]
The notation $^N\mathcal{P}_k$ is the permutation of $N$ taken $k$ at a time.

iii) $^{10}\mathcal{P}_6-^8\mathcal{P}_6$ That is the total possible minus the number of strings containing no symbols.

@
romsek, if it were posted in discrete mathematics I could move it. But for now tell Topquark

5. ## Re: A question related to Permuation and Combination

Originally Posted by Plato
The notation $^N\mathcal{P}_k$ is the permutation of $N$ taken $k$ at a time.

iii) $^{10}\mathcal{P}_6-^8\mathcal{P}_6$ That is the total possible minus the number of strings containing no symbols.

@
romsek, if it were posted in discrete mathematics I could move it. But for now tell Topquark
do you mean

$^{10}\mathcal{P}_6 - ^{7}\mathcal{P}_6$ ?

There are 3 symbol characters

(I see that you do, as it produces the correct answer)

6. ## Re: A question related to Permuation and Combination

Originally Posted by romsek
do you mean
$^{10}\mathcal{P}_6 - ^{7}\mathcal{P}_6$ ?
There are 3 symbol characters
(I see that you do, as it produces the correct answer)
Thank you, you are correct for some reason I did not count the $*$ as a symbol.

7. ## Re: A question related to Permuation and Combination

Hello, thanks for the solution .

But, I did this and i got a much lower value? 3 x 7 x 6 x 5 x 4 x 3 (one symbol) + 3 x 2 x 7 x 6 x 5 x 4 (two symbol) + 3 x 2 x 1 x 7 x 6 x 5 (three symbol)

8. ## Re: A question related to Permuation and Combination

Originally Posted by Zhen64
Hello, thanks for the solution .
But, I did this and i got a much lower value? 3 x 7 x 6 x 5 x 4 x 3 (one symbol) + 3 x 2 x 7 x 6 x 5 x 4 (two symbol) + 3 x 2 x 1 x 7 x 6 x 5 (three symbol)
Your calculation does give a lower number which is not correct.
You have counted the number of strings begin with one, two, or three symbols.

9. ## Re: A question related to Permuation and Combination

Oh okay, thanks !

10. ## Re: A question related to Permuation and Combination

Originally Posted by Zhen64
Oh okay, thanks !
Another way to do it is to choose the items you will use, then permute them. It would look like this:

Exactly 1 symbol: $\dbinom{3}{1}\dbinom{7}{5}$
Exactly 2 symbols: $\dbinom{3}{2}\dbinom{7}{4}$
Exactly 3 symbols: $\dbinom{3}{3}\dbinom{7}{3}$

Add that together and multiply by the number of permutations of the six chosen characters:

$\left[ \dbinom{3}{1}\dbinom{7}{5} + \dbinom{3}{2}\dbinom{7}{4} + \dbinom{3}{3}\dbinom{7}{3} \right] 6! = 146,160$

You can also use this for a similar formula to what Plato arrived at:

$\left[ \dbinom{10}{6} - \dbinom{7}{6} \right] 6! = 146,160$

Thanks alot.