Summary of permutations arrangements where the order counts r permutation from n different objects without repetition P n r n n r r n if n 0 1 1 n n n 1 if n 1 r permutation from n different objects with repetition P R n r n r k1 k 2 k n permutations of n different objects with limited repetition k1 k 2 k n k1 k 2 k n How many numbers from 1 1 1 2 2 3 can be constructed k1 6 Ans 3 2 k 2 k3 Combinations selections without reference to the order r combination from n different objects P n r C n r r Example 3 combinations from a b c d P 4 3 4 24 abc acd bcd bdc cab cba acd adc dca dac cad cda abd adb bad bda dab dba bcd bdc cdb cbd dcb dbc a b c a c d a b d b c d C 4 3 3 n P n r n C n r r n r r r r combinations of n objects without repetition The equivalence of 3 combinations from 4 objects and permutations of 4 objects with 3 of the same type a b c d 1 1 1 0 1 1 0 1 1 0 1 1 0 1 1 1 a b c a b d a c d b c d Combinations with repetitions Take for instance 4 combinations of a b a a a a a a a b a a b b a b b b b b b b We can consider this problem as the arrangements of 4 identical objects and one separator a a a a a a a b a a b b a b b b b b b b 5 permutations of 5 objects if 4 of the them are identical 5 4 Combinations with repetitions Donut shop has 5 types of donuts In how many ways we can select ten donuts This problem can be represented as an equivalent arrangement of ten donuts into 5 boxes All possible distributions Can be considered as permutations of a dozen of donuts and 4 separators between boxes One possible arrangement We need to count the number of permutations of 10 donuts and 4 separators So we have 14 objects 4 of which are identical and 10 are identical 14 C 14 4 10 4 From another side any arrangement can be viewed as a selection of 4 numbers out of 14 or 10 out of 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 The number of r combinations of n objects that can be repeated any number of times P n r 1 r C n r r R Can be considered as the number of arrangements of r identical objects and n 1 separators bars
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