A certain state has license plates showing three numbers and three letters. How many different license plates are possible
(a) if the numbers must come before the letters?
Answer:
for a license plate that has the order of numbers and letters set, we can just calculate the possible options for each position:
\[Number \times Number \times Number \times Letter \times Letter \times Letter\]
(10^3 * 26^3)## [1] 17576000\[10 \times 10 \times 10 \times 26 \times 26 \times 26 = 17,576,000\]
Now, if those positions can vary, we need to figoure out all the possible combinations of number and letter positions:
\[S = \left\{ NNNLLL, NNLLLN, .... \right\}\]
From our reading we know that it is easy to figure out the number of combinations using \(_{ n }{ C }_{ k }\) for this example which is \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right)\)
(combs <- choose(6, 3))## [1] 20(factorial(6)/(factorial(6 - 3) * factorial(3)))## [1] 20So, we can simply multiple the number of options found in (a) by the number of possible letter combinations:
(combs * 10^3 * 26^3)## [1] 351520000\[\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \times {10}^{3} \times {26}^{3} = 351,520,000\]
But, with license plates, order matters right? 123abc is not the same as 321cba. So we should really get all permuations, which is just \(_{ n }{ C }_{ k } \times k!\)
(choose(6, 3) * factorial(3))## [1] 120(120 * 10^3 * 26^3)## [1] 2109120000\[\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \times 3! \times {10}^{3} \times {26}^{3} = 2,109,120,000\]
Side note, initially I struggled with the idea of whether to pick numbers or letters (even though they were the same number), but check this out. If the question said there were four numbers and two letters, the amount of combinations would not change regardless of which we picked:
(choose(6, 4))## [1] 15(choose(6, 2))## [1] 15