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The first letter can be any one of 22. For each of these ...
The second letter can be any one of the remaining 21. For each of these ...
The third letter can be any one of the remaining 20.
So the number of different 3-letter line-ups is (22 x 21 x 20) = 9,240.
That's the answer if you care about the sequence of the letters, i.e. if you call ABC and ACB different.
If you don't care about the order of the 3 letters ... if ABC, ACB, BAC, BCA, CAB, and CBA are all
the same to you, then there are six ways to arrange each group of 3 different letters.
Then the total number of different picks is (9,240/6) = 1,540.
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How many different five-letter initials with none of the letters repeated can people have?
There are 26*25*24*23*22 = 7,893,600 sets of initials.
How many combinations can you make with 5 letters?
We first make an assumption that you are picking 5 letters from 26. If this is not the case, the problem is different. This is covered at the end of the answer. With repetition allowed (letters can be re-used), and order is not important since it is a combination (not a permutation), it is (26+5-1)!/(5!)(26-1)! This is a really big number! If it were a permutation with repetition, this is the same as the number of ways to pick 5 things from 26 with replacement. There are 26 choices for each letter, so you have 26x26x26x26x26 permutations. No repetition, order is important, it is 26! / (26-5)! = 26 x 25 x 24 x 23 x 22 = 7,893,600 No repetition, order doesn't matter (ABCDE is the same as EDCBA is the same as CEDAB, etc), it is 26! / (5!(26-5)!) = (26 x 25 x 24 x 23 x 22)/(5 x 4 x 3 x 2 x 1) = 65,780 What if you just have 5 letter already picked. Meaning you are not worried about picking the 5 from the alphabet, you already have 5 fixed letter. Some may be the same letter or not. If you allow for repetition of the letters, you have 5x5x5x5x5 permutations. For combinations, it would be 11!/5!4!=7983360 MOST LIKELY, you just have 5 distinct letters and you want to arrange them where order does matter and you cannnot repeat. In this case you have 5 choices for the first letter, 4 for the second, 3 for the third etc. So you have 5! or 120 choices. Of course if you have 5 things and the order does not matter, there is only 1 way to arrange them. For example. ABCDE is the same as ABEDC. So in dealing with combinations and permuations, the language must be quite exact. A small difference changes the entire problem. See related MathsIsFun link for more information.
How many cotton bolls in t-shirt?
22
You need to have a password with five letters followed by 2 odd digits between 0-9 inclusively If the characters and digits cannot be used more than once how many choices do you have?
(26 x 25 x 24 x 23 x 22) x (5 x 4) = 157,872,000
What is a estimate of 15 percent of 22?
15% of 22 = 15% * 22 = 0.15 * 22 = 3.3
