11/28/2023 0 Comments 13 permute 310 ways of placing the first duplicate, and for each of these, 3 ways of placing the second duplicate. Number of different duplicates for each letter = COMBIN(5,2) x COMBIN(3,2) = 30, i.e. Number of ways of selecting two duplicates, (i.e. Number of permutations with exactly two duplicates: No of ways of selecting remaining three X = 3! = 6 = PERMUT(3,3)ģ. No of different duplicates for each letter e.g. Possible duplicate letters = I,S or P = 3 Number of permutations with exactly one duplicate: Number of permutations with no duplicates = NIL obviously, plusĢ. Instead, you have to start thinking about the possible sub-sets, and determine the numbers and/or probabilities for each.įor example, if we want the number of 5-letter permutations you can make from MISSISSIPPI, you can add:ġ. In summary, all I'm saying is that once you get beyond the really simple questions, such as how many permutations of 3 can I make from 5 different letters, there is no simple, single formula that you can use manually, or codify into Excel. ![]() you have to think like a computer! I don't guarantee that my numbers above are correct, but the process is. So this approach is time consuming, and tricky. the number of permutations with 2 x s, 1 x X will be the same as for 2 x i, 1 x X. There will be some symmetries given that there are 4 i's and 4 s's, e.g. You'll need to consider many other combinations of duplicates, including So we need to eliminate 3 x 47 = 141 from the count. For each of these 3, there are 4x3=12 ways of selecting the i's and 4 ways of selecting the s, i.e. There are 3 ways this can happen: iis, isi or sii. Hence we need to eliminate 23 from the count. You can use a formula approach, but as I said in #7 above, it gets tricky listing all the duplicate possibilities to be eliminated. You can then use Data/Remove Duplicates to generate a list of non-duplicates, and count these.Ģ. This will list all 11x10x9 = 990 permutations, including duplicates. Use an algorithm like the VBA provided previously. If you want to count the number of permutations if you select three letters, say, you can do it two ways:ġ. ![]() Unfortunately there's no single simple formula when you are selecting less than the full number of letters. And similar for any other duplicated letter. With 4 i's always in the rearranged word, there are 4! ways of arranging i1, i2, i3 and i4, hence you need to divide by 4! to eliminate the i duplicates. There are 11! ways of re-arranging, including duplicates. ![]() and evaluate 6P 4.The 11 out of 11 case is relatively simple because it uses all the letters. With no repetitions, you can use the formula nP r = n(n - 1)(n - 2)(n - 3). How many four- letter passwords can be made using the six letters a, b, c, d, e, and f? Then, when the boys present, the number of arrangements is 6P 6 If the girls present first, then the number of arrangement is 8P 8 If the teacher is going to allow the girls to go first, how many different arrangement are there for the presentation? Word problem #5 6 boys and 8 girls will have a presentation in class today. ![]() The photographer has 1728 ways to arrange these people. If you have n choices for a first task and m choices for a second task, you have n × m choices for both tasks.
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