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  • What's the number of strings of digits and letters of length $7$ without repetition?

    • Solution 1. Pick $2$ digits, $4$ consonants, and $1$ vowel and then line them up: $\binom{10}{2}\binom{21}{4} \cdot 5 \cdot 7!$

    • Solution 2. Pick $2$ positions in the string for the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the spots: $\binom72\binom54 \cdot 10 \cdot 9 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 5.$

I think the methods above also count the permutations of MISSISSIPPI. But what about the problem below:

  • License plates in some state consist of $3$ different digits followed by $3$ different letters. How many of these plates are there?

    We choose places first and for each of those we choose symbols: $\binom63\binom33P(10, 3)P(26, 3).$ But apparently that's wrong. The actual answer is $P(10, 3)P(26, 3).$

What makes these two problems different? Why does choosing places first in the second problem results in wrong answer?

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For the pick of digits you have exactly 10 choices for first pick and 9 for the second and then 8 for the thrid giving a total of $$10\times9\times8=P(10,3)$$ similarly for the letters you have 26 choices for the first and 25 for the second and 24 for the third giving a total of $$26\times25\times24=P(26,3)$$ So the total number of license plates is $$P(10,3)\times P(26,3)$$

What makes it different is on a license plate the order does matter.

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