# Counting different kinds of permutations

• 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?

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 $$P(10,3)\times P(26,3)$$