Puzzle

Challenge 1

We asked : How many license plates could Seedonia’s MMV issue using six numbers?

The answer is relatively simple. With six numbers we can have any digit from 0 to 9 in any position on the license plate. There are ten choices for position 1. For every one of those ten choices, there are ten choices for position 2. So just with positions 1 and 2 there are 10 x 10 or 100 different combinations. For every one of those 100 combinations, there are ten more possibilities for position 3, or 1,000 combinations for the first three digits.

You get the idea. Every digit multiplies the number of possible combinations by 10. For a six-digit license plate there are 10⁶, or 1 million different combinations possible.

Here’s an even simpler way to think about it. Consider all the possible numbers from 000000 to 999999. That’s 1 million possibilities right there.

Challenge 2

In this challenge we wanted to know how many license plates the MMV could issue using two letters and four numbers on every plate?

This one’s a little harder. First we need to remember that the letters “I” and “O” are excluded. This leaves us 24 letters and ten numbers, or 34 possibilities for each digit.

If we use the same reasoning as before, we get 34 x 34 x 34 x 34 x 34 x 34 = 34⁶ possible combinations. 

Using scientific notation, 34 = 3.4 x 10¹, so 34⁶ = (3.4)⁶ x (10¹)⁶ = 1554.804416 x 10⁶ = 1.554804416 x 10³ x 10⁶ = 1.554804416 x 10⁹ = 1,554,804,416!

If you got this answer, you’re in good company, but there’s something you’re forgetting. There’s a flaw in this reasoning. Only two digits can be letters. In fact two digits must be letters and the other four must be numbers. So maybe there are 24 choices for the first two digits and 10 for each of the others. Using this reasoning we get

24 x 24 x 10 x 10 x 10 x 10 = 576 x 1000 = 5.76 x 10⁶ = 5,760,000.

This result is much less than our first try, but it still leaves something out. This is the number of combinations we’d get if the first two digits were letters and the last four were numbers. But the letters can appear anywhere among the six digits. We’ve got more work to do! We need to consider all possibilities for the placement of the two letters. Every one of them will have 5.76 x 10⁶ combinations.

Let’s give the positions names: P1, P2, P3, P4, P5, and P6. Now let’s figure out all the possible ways that two of them can be letters.

If P1 is a letter, we can have P1 - P2, P1 - P3, P1 - P4, P1 - P5, and P1 – P6, five possibilities.

IF P2 is the first letter, we can have P2 - P3, P2 - P4, P2 - P5, and P2 - P6, four possibilities.

IF P3 is the first letter, we can have three possibilities.

If P4 is the first letter, we have two possibilities.

If P5 is the first letter, we have only one possibility.

So all together we have 5 + 4 + 3 + 2 + 1 = 15 different placements for the two letters.

For each one, we have 5.76 x 10⁶ combinations of letters and numbers.

So the grand total of possible license plates is 15 x 5.76 x 10⁶ = 86.4 x 10⁶ combinations. That’s 86,400,000 different license plates, 86.4 times as many as we had using six digits.

Challenge 3

Finally, we asked if the population and the number of vehicles grow, which of the following alternatives would be better?

a) add one more number, for a total of seven, so each plate has two letters and five numbers

b) add one more letter, but keep the total at six, so each plate has three letters and three numbers

First let’s make a guess: adding an extra letter adds so many possibilities, so let’s guess that three letters and three numbers should give more combinations than two letters and five numbers. Now let’s see if we’re right.

It’s easy to figure out the result of adding another number: just multiply the total we got before by 10. So 86.4 x 10⁶ x 10 = 86.4 x 10⁷ = 8.64 x 10⁸ = 864,000,000. That’s almost 1 billion.

To get the number of possible combinations of three letters and three numbers, we multiply

24 x 24 x 24 x 10 x 10 x 10 = 13,824 x 10³ = 1.3824 x 10⁴ x 10³ = 1.3824 x 10⁷. That’s 13,824,000 combinations for each arrangement of three letters.

But how many arrangements are there? Let’s start making our lists again.

If P1 is a letter, we have [P1-P2-P3, P1-P2-P4, P1-P2-P5, P1-P2-P6], [P1-P3-P4, P1-P3-P5, P1-P3-P6], [P1-P4-P5, P1-P4-P6], [P1-P5-P6].

That’s 4 + 3 + 2 + 1 = 10.

If P2 is the first letter, we have [P2-P3-P4, P2-P3-P5, P2-P3-P6], [P2-P4-P5, P2-P4-P6], [P2-P5-P6].

That’s 3 + 2 + 1 = 6.

Can you see that if P3 is the first letter, we get 2 + 1 = 3 (possibilities), and if P4 is the first letter, we get only 1? (If you’re not convinced, work them out yourself.)

The total number of ways that three letters can be arranged in six positions is 10 + 6 + 3 + 1 = 20.

So our grand total is 20 x 1.3824 x 10⁷ = 27.648 x 10⁷ = 2.7648 x 10⁸ = 276,480,000, about 1 quarter billion.

Surprise: although the two values are close (what’s a few hundred billion here or there?), our guess was wrong!  We got only about one-third as many possible license plates with three letters and three numbers as we got using only two letters with five numbers, 8.64 x 10⁸ = 864,000,000.

What’s the moral of the story? In combinatorics it’s all in the details! So don’t be afraid when you come across a problem involving combinations. If you work carefully and use logic, considering all cases, you’ll figure it out!