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Problem:
Consider strings of digits and letters of length $7$ without repetition. Find the probability that a string contains $2$ digits, $4$ consonants, and $1$ vowel.
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Solution:
For the favorable outcomes, pick $2$ positions in the string for
the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the
spots: $\displaystyle{\frac{\binom 72 \times \binom 54\times 10 \times 9 \times 21 \times 20 \times 19 \times 18 \times 5}{36 \times 35 \times 34 \times 33 \times 32 \times 31 \times 30}}$ -
My question:
Counting the favorable outcomes part of the problem above looks like the MISSISSIPPI problem where the number of permutations is $\displaystyle{\binom{11}{4}\binom74\binom32\binom11}$. But in the case of MISSISSIPPI problem, after choosing the places in a string for the letters, we do not additionally fill them in as in the problem above. What's different between these problems? Thanks.
Question about the MISSISSIPPI problem
combinatoricsdiscrete mathematics
Best Answer
We'll assume that the "MISSISSIPPI problem" is determining how many permutations of the word "MISSISSIPPI" there are.
The idea is to think of the categories "digits, consonants, vowels" as the specific letters "M, I, S, P." For example:
In your problem, we choose where the digits will go, $\binom{7}{2}$, and then we multiply by the number of possible choices of distinct digits for those slots, $10 \times 9$.
In the MISSISSIPPI problem, we choose where the letter I's will go, $\binom{11}{4}$, but then there is no choice for what goes in those slots.