How many 5-digit permutations can be formed from this set: S=(1,2,2,3,3,3,4,4,4,4,5,5,5,5,5)?. Thanks for help.
+680 just multiply each case and divide by the repeating numbers
does that help?
it might not help if you don't know a lot about counting and probability.
+310 There are 15 numbers, and we are choosing 5 digits.
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Permutation means order matters, so there are 15 options for the first digit, 14 for the second, 13 for the third, etc.
15 14 13 12 11 = 15*14*13*12*11
But there are repeating digits in the set. We have to divide by the number of ways the 2s, 3s, 4s, and 5s can repeat. Since there are two repeating 2s, there are 2! different ways they can repeat. Since there are three repeating 3s there are 3! different ways they can repeat (using the same logic as when we did 15 options, 14 options, 13 options, etc.
(15*14*13*12*11) / (2!*3!*4!*5!)
However, this doesn't come out as an integer :( ... I'm not sure what I did wrong. Maybe you can find my mistake. Hopefully this helps anyways!
It is generally much easier to compute the "combinations" first and then convert the combination into permutations. With a very short computer code, each of these combinations is converted to permutations and then summed up. I wrote such a short computer code just for the pupose of converting these 71 combinations into permutations and the result is: 2,111 permutations.
12233 , 12234 , 12235 , 12244 , 12245 , 12255 , 12333 , 12334 , 12335 , 12344 , 12345 , 12355 , 12444 , 12445 , 12455 , 12555 , 13334 , 13335 , 13344 , 13345 , 13355 , 13444 , 13445 , 13455 , 13555 , 14444 , 14445 , 14455 , 14555 , 15555 , 22333 , 22334 , 22335 , 22344 , 22345 , 22355 , 22444 , 22445 , 22455 , 22555 , 23334 , 23335 , 23344 , 23345 , 23355 , 23444 , 23445 , 23455 , 23555 , 24444 , 24445 , 24455 , 24555 , 25555 , 33344 , 33345 , 33355 , 33444 , 33445 , 33455 , 33555 , 34444 , 34445 , 34455 , 34555 , 35555 , 44445 , 44455 , 44555 , 45555 , 55555 , Total = 71 combinations.
