This gives us 916132832 permutations for a password! **For future reference, it is good to note that if the problem said "each letter, lowercase letter, or capital letter can only be used once". then the answer would be 62 * 61 * 60 * 59 * 58.
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Similarly we have 62 combinations for each position. Therefore total number of combinations are=62 * 62 * 62* 62 *62 =916132832 Answer = Total number of combination - combination which does not follow 3rd condition I am not able to figure out how to calculate number of combinations which does not follow 3rd condition.- For a 5-character password, there are 62 x 62 x 62 x 62 x 62 = 916132832 possible unique passwords. Therefore, the total possible unique passwords is 238328 + 14776336 + 916132832 = 783264. 7. To solve this, we can use the formula for permutations with repetition. This formula is n^r, where n is the number of items and r is the number of slots.
- Power of 62 Table 1 to 10 62 1 = 62 62 2 = 3844 62 3 = 238328 62 4 = 14776336 62 5 = 916132832 62 6 = 56800235584 62 7 = 3521614606208 62 8 = 218340105584896 62 9 = 13537086546263552 62 10 = 839299365868340200
- What is 62 to power of 5? The answer is 916132832. Concepts of base and exponent, how to calculate powers, animated solultion steps, instructions.
- There are 916132832 5 character usernames There are 56800235584 6 character usernames (enough for everyone in the world to have 8 usernames each) This pattern continues. Although most of these don't make sense, we won't run out of usernames in the foreseeable future. And as usernames get longer, the number available goes up very quickly.
ikuro's-homepage
https://ikuro-kotaro.sakura.ne.jp › koramu2 › 9210_d9.htm
（その11）の続き． （60）^5＝777600000 （61）^5＝844596301 （62）^5＝916132832 （63）^5＝992436543 （64）^5＝1073741824 （65）^5＝1160290625 ...ikuro's-homepage
https://ikuro-kotaro.sakura.ne.jp › koramu2 › 42477_q6.htm
（その12）の続き． （60）^5＝777600000 （61）^5＝844596301 （62）^5＝916132832 （63）^5＝992436543 （64）^5＝1073741824 （65）^5＝1160290625 ...
- How to find number of combinationshttp://math.stackexchange.com/questions/3176538/ddg#3176554
$26$ ways to choose the capital letter and $5$ choices for its place.
$10$ ways to choose the digit and $4$ places to put it.Concerning the remaining three places where to fit small letters, you shall distinguish among:
- three equal letters: $26$ choices, $1$ way to arrange
- two equal letters: $26 \times 25$ choices, $3$ different arrangements
- all different letters: $26 \times 25 \times 24$ ways.total: $\quad26 \cdot 5 \cdot 10 \cdot 4 \cdot 26 \cdot\left( 1+25 \cdot 3+25 \cdot24 \right)=91 395 200$
--G Cab
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