How many 3-letter combinations are possible from A, E, I, O, and U? Show work please!?
Okay, the letters A E I O U can be place in these three places: _ _ _. There are 5 possibilities for the first blank: - _ _. Once that one is filled, there are four letters left to fill the second blank: - - _. Once both of the first two spots are full, then there are 3 letter choices for the last blank: - - -. To find out the number of possibilities, multiply the first possible number of choices times the second times the third, so: 5 * 4 * 3 = 60 possibilities. There are four cities to choose from and three reports to do. So, it's the same as above. There are three places: _ _ _. When the first is full, there are 3 cities left: - _ _. Then the first and second are full, there are 2 cities left: - - _. Then there are only two options to fill that third space. So 4 * 3 * 2 = 24 possibilities.
How many 2 letter combinations are there?
26*26 = 676 super combos
How many 2 letter combinations are there?
like: aa ab ac except with every letter and every combination. How many 2 letter combinations are there? I need to find out for my math class and I don't know how, and/or is there a website that shows them ALL? Ok thnks :)
How many 2-letter combinations exist in the organized list of 2-letter permutations shown below?
Hi, There are 6 two-letter combinations of the letters A, B, and C. <==ANSWER If you are talking about putting together two different pairs of these pairs of letters, like AB-CA or AB-AC, then there are 30 combinations. I hope that helps!! :-)
How can I create a list of all 3 letter combinations from the alphabet with excel sorted in a specific way?
It takes about a minute to finish, but i guess you are seeking the result. Sub a2zRpt() Application.ScreenUpdating = False Dim fstArr Dim checkAv As Boolean checkAv = False count = 1 For j = 1 To 25 c = -1 For i = 1 To 25 If Chr(96 + j) >= Chr(96 + i + 1) Then c = c + 1 Else c = i End If Cells(count, 1) = Chr(96 + j) & Chr(96 + j) & Chr(96 + c + 1) count = count + 1 Cells(count, 1) = Chr(96 + j) & Chr(96 + c + 1) & Chr(96 + j) count = count + 1 Cells(count, 1) = Chr(96 + c + 1) & Chr(96 + j) & Chr(96 + j) count = count + 1 Next Next ReDim fstArr(Cells(1, 1).End(xlDown).Row) fstArr = Range("A1:A" & Cells(1, 1).End(xlDown).Row).Value For i = 1 To 26 For j = 1 To 26 For k = 1 To 26 temp = Chr(96 + i) & Chr(96 + j) & Chr(96 + k) For h = 1 To UBound(fstArr) temp = Chr(96 + i) & Chr(96 + j) & Chr(96 + k) If temp = fstArr(h, 1) Then checkAv = True Exit For Else checkAv = False End If Next If Not checkAv Then Cells(count, 1) = Chr(96 + i) & Chr(96 + j) & Chr(96 + k) count = count + 1 hhh = False End If Next Next Next End Sub
What can be added to this list of non-English letter combinations?
I don't know how you plan to use this list, but I assume you're looking for a minimal set, so for example if you have BB in the list you can omit BBA, BBB, and so on.First you really need to specify a maximum length of the combinations in your list. Otherwise for every word in the dictionary that isn't also the start of a longer word, you're going to have to add 26 new combinations, e.g. RHUBARBA, RHUBARBB, RHUBARBC and so on. Even with a maximum length of 3 letters, I'd guess you're looking at a list of a few thousand combinations. (There are 26x26x26=17576 possible 3-letter combinations, most of which don't occur at the start of any words, but we would skip the combinations in which the first TWO letters don't start any words either).I'll assume a maximum length of 3 letters.For each letter of the alphabet, go through the dictionary in order and:- for each letter which does not occur as the second letter of any words starting with the first letter, add that two-letter pair to your list.- for each letter which DOES appear after the second letter in any word, go through all possible third letters in order, and add all three-letter combinations which do not occur at the start of any words.Obviously repeat the same process if you want longer combinations in your list.I don't even know if this is what you intended by the question, but there are six people waiting for an answer, so I hope this is useful to SOMEONE...!
Java program to make letter combinations?
I know all of that.... This is what I have so far... String word = "a"; do{ word+="a"; for(int x = 0; x< word.length;x++) { do{ //Go to website ("http://www."+word+".com"); word.charAt(x)++; }while (word.charAt(x)! = 122); } }while (true);
How can I generate a spreadsheet with all letter combinations for all 1, 2 and 3 letter words?
I assume that you want all letter combinations that comprise 1, 2 or 3 letters. If you want to limit yourself to valid words, you'll need a dictionary file. If you have a dictionary file, you don't need any formulas, just cut/paste the dictionary. It is not that long.But if what you really want is a listing of all letter combinations, create a formula that emits all the letter combinations. You can create it out of combinations of formulas. First, make a formula that converts a number to a letter. That number can be the current row number or you can have a column of numbers. That is easy. Now create a formula that, for a given number, divides by 26 and tells you the integer portion, and another that tells you the remainder. Do the same for a number that does that for 26*26 and for 26. Then, create an If() statement that uses one formula if the number is between 1 and 26, another formula if the number is between 27 and 26*27, and another formula if the number is between 26*27+1 and 26*26*27. That last formula will tell you how many letters to output.
How many different ways can you make combinations of 3 letters out of the five-letter word LEARN?
When you use the word combination then order does not matter, and LEA and ALE are considered the same. In this case you have 5 letters to choose for the 1st spot, 4 choices for the second spot and 3 choices for the 3rd spot, or 5x4x3=60, but because LEA and ALE are considered the same you must divide by how many ways you can arrange 3 letters, 3 choices for the 1st, 2 choices for the 2nd and 1 choice for the last or 3x2x1=6. So 60/6 = 10.If on the other hand you consider ALE and LEA to be different then you are talking permutations not combinations, and the math is actually much easier because it is just 5x4x3=60.