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A password shall begin with 3 different capital letters, followed by 3 to 5 digits between 0 to 9.

a)How many possibilities of passwords are there?

b)How many passwords with the character string "B4" are there?

(with binomial Coefficient = $n \choose k$)

a)For the 3 first digits I have $26\choose 3$ possibilities. Then they are followed by 3 to 5 digits, and I don't know how to approach...

3 Answers 3

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a) There are 3 different cases for this. For when there are 3 numbers, 4 numbers, 5 numbers

There are 3 different letters with 26 to choose from. Hence $\displaystyle\binom{26}{3}$ but this doesn't take into account order so we multiply by $3!$. Then for case 3, 4, 5. Since there are 10 numbers to choose from. We multiply $\displaystyle\binom{26}{3}\times 3!$ by $10^3$, $10^4$, $10^5$ respectively for cases when we use 3, 4, 5 digits. I assumed here that the numbers can be repeated.

b) Since the character string is "B4" (a letter then a number) then B is in the 3rd position, and 4 is in the 4th position.

$_,_,B,4,_,_$ is our format. With permutations $25\times24\times1\times1\times10\times10$

Since B and 4 are fixed we have 1 for the 3rd and 5th position. And $25, 24$ since we alread chose $B$ for the 3rd character.

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    $\binom{26}{3}\cdot 3!$ seems an awefully roundabout way of writing $26\cdot 25\cdot 24$. But then again, the question was asked with binomial coefficients in mind.2012-11-22
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    So order means, that every number or letter has its assigned place/digit?2012-11-22
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    @user49421 Yes. The binomial coefficients are used only when you don't care which is first, which is second and so on, you just care which are picked and which aren't. Of course, you can use binomial coefficients and then take care of ordering afterwards, as is done in the answer above.2012-11-22
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    Yes. I actually shouldv'e just written $26P3$ Lol. :) But it was just to clarify the point of order. Yes. For each permutation of letters and numbers, every letter has a designated position. For example if we want to put order, ABC is different from BAC. But if we don't consider order, ABC would be the same as BAC. But since we are taking passwords into consideration. We should need to take into account order to make guessing more "complex" Haha :)2012-11-22
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    Now I have understood a) :), and because of this order of place I also have to write 10^k with k=place/digit?2012-11-22
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    We need to multiply $26\times25\times24$ by $10^k$ where $k \in \{3,4,5\}$ because there are 10 numbers to choose from (with repetition). We select a number by $3,4,5$ times. I'll just give the case where we select 3 numbers. Suppose we select the string "123" this is different from the string "234" for each place of the digits we can choose 10 numbers. But there are 3 places. Hence we multiple 10 a number of 3 times to create all possibilities for our 3 number string.2012-11-22
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    yes, and why can't I calculate b), this way 26x25x(10^2+10^3+10^4)?or is it because the 26 letter is already used by B?2012-11-22
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    Yes. For b) you want your letters to be distinct right? Since you already have 1 letter (which is B), we only have 25 choices left for the first position. And since we chose the first position, the 2nd position has 24 choices left. And it should be $(10^3 + 10^4 + 10^5)$ not $(10^2 + 10^3 + 10^4$, remember that you choose 3,4, or 5 numbers and not 2,3, or 42012-11-22
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    but what's about the 4. digit with the 4, which is already used, is it not therefore that (3 to 5)-1 => (10^2+10^3+10^4) instead of (10^3+10^4+10^5) ?2012-11-22
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    Ahh! I overlooked that one. Yes, That is correct. See what I got for the 3 digit case in my solution. That is the part of the $10^2$ in your sum. :)2012-11-22
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    Ok so the solution is 25*24*(1*1)*(10^2+10^3+10^4).I believe after all these helpful comments, I have a much better understanding of the topic.Thank you :)2012-11-22
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a) You can pick the first letter in 26 ways (1 out of 26 letters). The next one will be from the remaining 25 letters, so ways of picking it would be 25. Similarly, the 3rd letter can be picked in 24 ways. So picking the first 3 letters can be done in 26x25x24 ways. For the numbers, ways of picking each digit can be done in 10 ways. As having 3 digits, 4 digits and 5 digits are mutually exclusive events, we can add the ways of generating each. Thus, for 3 digits, we have 26x25x24x$10^3$ ways. Similarly for 4 and 5 digits, giving the total number of ways to be: 26x25x24x($10^3 + 10^4 + 10^5$).

b) You need to fix the 3rd char to B and the first digit to 4 (that is the only place where a number follows a letter). Select the rest from the remaining 25 letters like above, to give the number of ways as: 25x24x($10^2 + 10^3 + 10^4$).

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    For b) we already chose 1 letter. So there are 25 letters to choose from for the first position. I think. =)2012-11-22
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    @KeithReyes Yep, you are right. Thanks! Fixed.2012-11-22
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a) How about ${10}^3+{10}^4+{10}^5$? (EDIT: Sorry, I first had assumed the digits shold be different as well)

b) Hint: Where can "B4" occur at all?

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    that means I only have to add from the third to the 5th digit this sum.Ok and for b) it is only one possibility, because it's the only one where a letter is followed by a digit?2012-11-22
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    But in a) didn't take into account the permutations for the letters. The sum should be multiplied by $26\times25\times24$2012-11-22