Combinations and permutations

\(\text{STATISTICS = SSSTTTIIAC}\\ \text{So in forming distinct permutations we must consider the various cases }\\ \text{of how many pairs or triples are in a given choice of 4 letters}\\\)

\(\text{case 1: 1 triplet, 1 singleton}\\ \text{There are }\dbinom{2}{1}=2 \text{ ways to select the triplet,}\\ \dbinom{4}{1}=4 \text{ ways to select the singleton, }\\ \text{and }\dbinom{4}{1}=4 \text{ ways to select where the singleton falls}\\ n_1 = 2\cdot 4\cdot 4 = 32\)

\(\text{case 2: 2 pairs}\\ \dbinom{3}{2}=3 \text{ ways to select which letters form the two pairs}\\ \dbinom{4}{2}=6 \text{ ways to select which slots for the first pair}\\ n_2 = 3\cdot 6 = 18\)

\(\text{case 3: 1 pair}\\ \dbinom{3}{1}=3 \text{ ways to select which letter is the pair}\\ \dbinom{4}{2} = 6 \text{ ways to select the singleton letters}\\ \dbinom{4}{2} = 6 \text{ ways to assign the pair of letters}\\ \dbinom{2}{1}=2 \text{ ways to assign the first singleton}\\ n_3 = 216\)

\(\text{case 4: 4 singletons}\\ \dbinom{5}{4}=5 \text{ ways to select the letters}\\ 4!=24 \text{ ways of arranging them }\\ n_4 = 120\)

\(n_p = 32+18+216+120 = 386\)

\(\text{Combinations is much easier. }\\ \text{We still have to consider cases}\\ \text{case 1: 1 triplet }c_1=\dbinom{2}{1}\dbinom{4}{1} = 8\)

\(\text{case 2: }c_2=\dbinom{3}{2}=3\)

\(\text{case 3: }c_3 = \dbinom{3}{1}\dbinom{4}{2} = 18\)

\(\text{case 4: }c_4=\dbinom{5}{4} =5\)

\(n_c = 8+3+18+5 = 34\)

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.

Thank you very much Rom. I greatly appreciate your time. However, the answers given to us are:

Permutations = 386,  Combinations = 125. Both of them are supposed to be "Distinct". But, as of yet, no detailed explanations as to how they were arrived at!. They are also listed in some sort of "lexicographic" order, which I have.

Thanks Rom. Here are the "combinations" listed:

{s, t, a, t} | {s, t, a, i} | {s, t, a, s} | {s, t, a, c} | {s, t, t, i} | {s, t, t, s} | {s, t, t, t} | {s, t, t, c} | {s, t, i, s} | {s, t, i, t} | {s, t, i, i} | {s, t, i, c} | {s, t, s, t} | {s, t, s, i} | {s, t, s, c} | {s, t, s, s} | {s, t, c, s} | {s, a, t, i} | {s, a, t, s} | {s, a, t, t} | {s, a, t, c} | {s, a, i, s} | {s, a, i, t} | {s, a, i, i} | {s, a, i, c} | {s, a, s, t} | {s, a, s, i} | {s, a, s, c} | {s, a, s, s} | {s, a, c, s} | {s, i, s, t} | {s, i, s, i} | {s, i, s, c} | {s, i, s, s} | {s, i, t, i} | {s, i, t, c} | {s, i, t, s} | {s, i, i, c} | {s, i, i, s} | {s, i, c, s} | {s, s, t, i} | {s, s, t, c} | {s, s, t, s} | {s, s, i, c} | {s, s, i, s} | {s, s, c, s} | {t, a, t, i} | {t, a, t, s} | {t, a, t, t} | {t, a, t, c} | {t, a, i, s} | {t, a, i, t} | {t, a, i, i} | {t, a, i, c} | {t, a, s, t} | {t, a, s, i} | {t, a, s, c} | {t, a, s, s} | {t, a, c, s} | {t, t, i, s} | {t, t, i, t} | {t, t, i, i} | {t, t, i, c} | {t, t, s, t} | {t, t, s, i} | {t, t, s, c} | {t, t, s, s} | {t, t, t, i} | {t, t, t, c} | {t, t, t, s} | {t, t, c, s} | {t, i, s, t} | {t, i, s, i} | {t, i, s, c} | {t, i, s, s} | {t, i, t, i} | {t, i, t, c} | {t, i, t, s} | {t, i, i, c} | {t, i, i, s} | {t, i, c, s} | {t, s, t, i} | {t, s, t, c} | {t, s, t, s} | {t, s, i, c} | {t, s, i, s} | {t, s, c, s} | {a, t, i, s} | {a, t, i, t} | {a, t, i, i} | {a, t, i, c} | {a, t, s, t} | {a, t, s, i} | {a, t, s, c} | {a, t, s, s} | {a, t, t, i} | {a, t, t, c} | {a, t, t, s} | {a, t, c, s} | {a, i, s, t} | {a, i, s, i} | {a, i, s, c} | {a, i, s, s} | {a, i, t, i} | {a, i, t, c} | {a, i, t, s} | {a, i, i, c} | {a, i, i, s} | {a, i, c, s} | {a, s, t, i} | {a, s, t, c} | {a, s, t, s} | {a, s, i, c} | {a, s, i, s} | {a, s, c, s} | {i, s, t, i} | {i, s, t, c} | {i, s, t, s} | {i, s, i, c} | {i, s, i, s} | {i, s, c, s} | {i, t, i, c} | {i, t, i, s} | {i, t, c, s} | {i, i, c, s}=125 

I can offer a guess from where this lexicographic order of BS originated. A mathematics professor usually creates this type of teaching material and he or she will recruit an undergrad (or graduate) student to provide solutions with answers as a paid assignment. The students, and the professors who recruit them, are notorious for not checking and cross checking their solutions for accuracy. (This is why the answers in the “back of the book” are often wrong.)

Obviously, this lexicographic output was generated by a miscoded computerized algorithm that cataloged the correct set along with a massive amount of rubbish. Even if correctly presented, computerized solution algorithms of this type are of limited use beyond verifying a solution.

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Along with Rom’s use of binomial counting for each case, there are two other methods for solving this:  A Markov matrix and a generating function. 

Though sometimes difficult to create, the generating function is relatively simple for this application.  For “SSSTTTIIAC” the counts are 3, 3, 2, 1, and 1 for each of the letters. This corresponds to a generating function of (1+x+x^3) (1+x+x^3) (1+x+x^2)(1+x)(1+x)

Expanding this series produces the coefficients.  (Doing this manually takes about 45 minutes. The “Wolf” takes about 10 seconds.)

\(x^{10} + 5 x^9 + 13 x^8 + 24 x^7 + 34 x^6 + 38 x^5 + \mathbf {34 x^4} + 24 x^3 + 13 x^2 + 5 x + 1\\\)

Here, the coefficient of 34 for x⁴ corresponds to the 34 unique-element subsets of four (4) from a set of 10 elements (S,S,S,T,T,T,I,I,A,C). Note the inclusion of all coefficients for subsets ranging from zero (0) to ten (10).

GA