[Math] How many different Strings can I create with the characters of the String SPEISESTAERKE

combinatoricsdiscrete mathematics

How many different Strings can I create with the characters of the String SPEISESTAERKE ?

According to that post, I have tried the following. Note, that I use all the given characters at once to create all possible combinations:

$2 \times$S,
$1 \times$P,
$4 \times$E,
$1 \times$I,
$1 \times$A,
$1 \times$R,
$1 \times$K

$\binom {13}{2} \cdot \binom {11}{1} \cdot \binom {10}{4} \cdot \binom {6}{1} \cdot \binom {5}{1} \cdot \binom {4}{1} \cdot \binom {3}{1} \cdot \binom {2}{1} \cdot \binom {1}{1} = 129 729 600$

Unfortunately this is the wrong answer. The correct solution is $432 432 00$.

Question: How can I achieve the correct result?

Best Answer

We can calculate the easiest case first and then tailor the number for the purpose. We pretend the letters are all different; then, since there are 13 letters, we can obtain $13!$ words. But there are replicate ones. The "S" appears three times and the "E" appears four times. So $13!$ is $3! \times 4!$ times the number required! So the required number is $$ \frac{13!}{3!4!}. $$

Related Solutions

[Math] How many possible combinations of a HEX string

I think the sanest way to split this is by how many different letters and numbers is contained in the string. Call the total number of different symbols $k$. The possibilities are then:

k=5    1+4  2+3  3+2  4+1
k=6    1+5  2+4  3+3  4+2  5+1
k=7    1+6  2+5  3+4  4+3  5+2  6+1
k=8    1+7  2+6  3+5  4+4  5+3  6+2
k=9    1+8  2+7  3+6  4+5  5+4  6+3
k=10   1+9  2+8  3+7  4+6  5+5  6+4

For each of the 33 entries in the table, the number of ways to select which hex digit appear at all is a simple product of binomial coefficients.

From there each row of the table can be summed and treated uniformly. First select which $10-k$ of the $k$ symbols are going to appear twice, for a factor of $\binom{k}{10-k}$. Then multiply by $10!$ for the ways to arrange the symbols you have selected, and divide by $2^{10-k}$ to account for the fact that each pair of two equal symbols cannot be distinguished.

[Math] How many different $4$ letter words can you create from the characters of “parallelogram”

The following is a little ugly. Note that we have $3$ l's, $3$ a's, 2 r's, and $5$ singleton letters.

Divide into cases:

(i) If we will use $3$ l's, their locations can be chosen in $\binom{4}{1}$ ways, and the remaining letter can be chosen in $\binom{7}{1}$ ways, for a total of $\binom{4}{1}\binom{7}{1}$.

There are also $\binom{4}{1}\binom{7}{1}$ words with $3$ a's.

Or else we could have said the tripled letter can be chosen in $\binom{2}{1}$ ways. for each choice, where it goes can be chosen in $\binom{4}{3}$ ways, and then the remaining slot can be filled in $\dots$.

(ii) If the word will have $2$ doubled letters, the types of the letters can be chosen in $\binom{3}{2}$ ways. For each of these ways, the places where the earlier (in the alphabet) of the chosen letters go can be chosen in $\binom{4}{2}$ ways.

(iii) If the word will have exactly $1$ doubled letter, that letter can be chosen in $\binom{3}{1}$ ways. Continue.

(iv) If the word is to have no repeated letter, we have a simple situation, an alphabet of $8$ letters.

Best Answer

Related Solutions

[Math] How many possible combinations of a HEX string

[Math] How many different $4$ letter words can you create from the characters of “parallelogram”

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