Solution: Show
Five letters can be chosen from the 26 alphabet letters in the following combinations. \(^{26}{C_5}\) = 26!/(21! 5!) = (26 × 25 × 24 × 23 × 22 × 21!)/(21! 5!) = (26 × 25 × 24 × 23 × 22)/120 = (26 × 5 × 23 × 22) = 65,780 ways The five letters selected in each group can be arranged in 5! Ways or 120 ways. Therefore the different arrangements of five letters that can be made from 26 letters of the alphabet are 65,780 × 5! ways = 65,780 × 120 = 7,893,600 ways How many different arrangements of 5 letters can be made from the 26 letters of the alphabet?Summary: The different arrangements of five letters that can be made from 26 letters of the alphabet are 7,893,600 ways. There are 26! (which means 26*25*24 all the way to 1) permutations. this comes to 403,291,461,100,000,000,000,000,000 different combinations. If it takes you thirty seconds to write out one combination, then it would take 383,123,823,100,000,000,000 YEARS to write all the combinations. To put this into context, the universe (and possibly time itself) is only 13,700,000,000 years old. Just in case you hadn't seen the answer to this it is 26!, Also known as 26 factorial His combination of ABC having 6 combinations is because to workout the combinations with 3 separate letters, ABC, you work out 3!, Or 3 factorial And factorial is multiplication of all numbers between 1 up to and including the given integer So for ABC, 3! 3 * 2 * 1 = 6 For A-Z, it is 26! 26 * 25 * 24 * 23 * (all the way to 1)... 4 * 3 * 2 * 1 = 8 * 10²⁵ ish.
Anthony J. Each letter or number has the same probability of appearing in each character in x length. What is the probability of "abcdefghij" (a 10 letter string) to appear if each letter and number in there had the same probability to appear. What is the forumla to determine this. The only changing factor is x, the length of string that generates. Originally, I had this... What was I trying to calculate with this? (26 + 10) * x 2 Answers By Expert Tutors
Hamilton A. answered • 12/28/15 Degrees in Math, 10+ Years Tutoring What @Christine said is true: the probability of seeing a particular string of length x, in a string of length x that consists of random letters and digits, is 1 / (36 ^ x). This is true because, at each position in the string, there are 36 possible characters that could go there, so there will be 36 * ... * 36 = 36^x different strings, all equally likely, and we're just interested in one particular string. But that doesn't fully answer the question you asked, which was, "What's the probability of seeing 'abcdefghij' in this random string?" Clearly, if x < 10, then the probability of seeing 'abcdefghij' is zero, since you can't have a length-10 string appear in fewer than 10 characters. If x = 10, then we're in the case discussed previously: the string we're interested in has the same length as the number of characters available, so the probability is 1 / 36^10. If x > 10, then it gets a little bit trickier, since our magic substring 'abcdefghij' could start at any position where it has enough room. For example, with x = 11, it could start at either position 1 (and take up characters 1-10) or position 2 (and take up characters 2-11). With x = 12, it could start at positions 1, 2, or 3. In general, for x > 10, there are (x - 10 + 1) positions where the magic substring could start. Let's consider the case x = 12 for a minute. The probability that the magic length-10 substring starts at a given position is still 1 / 36^10, and any character is allowed in the other positions, so for example, P(magic substring starts at position 1 in a length 12 string) = (1 / 36^10) * 1 * 1, where the 1s represent the fact that any character can appear in positions 11 and 12. The event "magic substring occurs in a length-12 string", which is what we want the probability of, is the union of three events:
These are disjoint events (since, for example, the magic substring starting at position 1 means that "b" is in position 2, so "abcdefghij" can't start in position 2), and so the probability of the union, i.e. the probability of seeing our length-10 magic substring in a length-12 string of random characters, is the sum of the probabilities of those three events: 3 / 36^10. For better or worse, this reasoning only holds while our magic substring has length < 2*x. Why? Because these events are no longer disjoint when you have at least twice as many characters in your random string as you do in the substring you're looking for. For example, if you have 20 or more random characters to work with, 'abcdefghij' can appear multiple times. But this becomes a pretty hard problem, and probably wasn't what your teacher was looking for...
 Christine O. answered • 12/27/15 UVA Student tutoring in Computer Science, Math, and SAT prep (26 + 10)^x would give you the total number of possible combinations that could be created by arranging 26 letters and 10 numbers in a string of x length. Therefore, the probability of encountering one specific string of x length would simply be: 1 / ((26 + 10)^x ) Still looking for help? Get the right answer, fast.ORFind an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. How many ways can 5 letters be arranged 26?The different arrangements of five letters that can be made from 26 letters of the alphabet are 7,893,600 ways.
How many combinations of 26 letters are there?Answer and Explanation:
The number of possible combinations that are possible with 26 letters, with no repetition, is 67,108,863.
How many combinations can be made with 26 letters and 10 numbers?Therefore there are 3636 combinations, each of length 36 that can be made from 26 letters A-Z and 10 numbers 0-9.
How many 5 digit number and letter combinations are there?10 * 9 * 8 * 7 * 6 = 30,240 possible codes.
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How many combinations are there with 26 letters and 5 numbers?
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