\(\text{STATISTICS = SSSTTTIIAC}\\ \text{So in forming distinct permutations we must consider the various cases }\\ \text{of how many pairs or triples are in a given choice of 4 letters}\\\)
\(\text{case 1: 1 triplet, 1 singleton}\\ \text{There are }\dbinom{2}{1}=2 \text{ ways to select the triplet,}\\ \dbinom{4}{1}=4 \text{ ways to select the singleton, }\\ \text{and }\dbinom{4}{1}=4 \text{ ways to select where the singleton falls}\\ n_1 = 2\cdot 4\cdot 4 = 32\)
\(\text{case 2: 2 pairs}\\ \dbinom{3}{2}=3 \text{ ways to select which letters form the two pairs}\\ \dbinom{4}{2}=6 \text{ ways to select which slots for the first pair}\\ n_2 = 3\cdot 6 = 18\)
\(\text{case 3: 1 pair}\\ \dbinom{3}{1}=3 \text{ ways to select which letter is the pair}\\ \dbinom{4}{2} = 6 \text{ ways to select the singleton letters}\\ \dbinom{4}{2} = 6 \text{ ways to assign the pair of letters}\\ \dbinom{2}{1}=2 \text{ ways to assign the first singleton}\\ n_3 = 216\)
\(\text{case 4: 4 singletons}\\ \dbinom{5}{4}=5 \text{ ways to select the letters}\\ 4!=24 \text{ ways of arranging them }\\ n_4 = 120\)
\(n_p = 32+18+216+120 = 386\)
\(\text{Combinations is much easier. }\\ \text{We still have to consider cases}\\ \text{case 1: 1 triplet }c_1=\dbinom{2}{1}\dbinom{4}{1} = 8\)
\(\text{case 2: }c_2=\dbinom{3}{2}=3\)
\(\text{case 3: }c_3 = \dbinom{3}{1}\dbinom{4}{2} = 18\)
\(\text{case 4: }c_4=\dbinom{5}{4} =5\)
\(n_c = 8+3+18+5 = 34\)
Note that you can separate the permutations calculations into factors related to selecting the letters
and factors related to arranging the letters.
Combination calculations just omit the factors related to arranging the letters.
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