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A square of size 4 gives you an area and perimeter of 16.
A rectangle with sides of 3 and 6 gives you an area and perimeter of 18.
So halfway between 16 and 18 is 17. Is that what you're asking?
You have 26 * 25 * 24 combinations of letters you can use assuming no duplicates. That's 15600. You can choose from 1 to 9 for the 1st digit. For the second digit you have 9 choices also; 0 through 9 except for whatever was picked for the 1st digit. The 3 digit you have 8 choices. So 9*9*8 possibilities for the numeric part of your plate. That's 648. Multiply those two numbers together: 648 * 15600 = 10,108,800.
I'm not sure what you mean by "letters and numbers must be combined". Do you mean the letters must be grouped together and the numbers must be grouped together? ABC123 is valid but AB123C is not? If so, then the only 2 formats possible are AAA### and ###AAA. In that case multiply the 10108800 by 2 for the total possibilities.
The reason that I brought up the whole Base 3 representation is to show it is just a "limitation" of our decimal system. Certain numbers, like 1/3 can't be expressed 100% accurately in the decimal system. But it can be expressed exactly in base 3. .33333... (base 10) = 1/3 = .1 (base 3). So for cases like these, switching to a different base can be helpful.
You could also use a system other than decimal.
3.333.. = 3 1/3 (base 10) = 10.1 (base 3)
6.666.. = 6 2/3 (base 10) = 20.2 (base 3)
In base 3:
10.1
20.2
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101.0
101 (base 3) = 10 (base 10).
Q2 When 3 sqaures are drawn so that sides are shared there are two figures which can be constructed.
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You're right on the 6 letter words.
Let's do 5 letter words first. You're concerned about 2 groups of letters; those groups without duplicates and those with duplicates (the "a"). There is only one group of 5 letters without duplicates. You can arrange those 5! different ways or 120.
Now let's consider the 5 letter groups with 2 a's. You need 3 more letters from the remaining four, i.e. (4 choose 3) which is 4. You can arrange those groups of 5 letters in 5! /2! different ways (60).
So the number of 5 letter words is 120 + (4 * 60) = 360.
4 letter words:
Number of words without duplicates: Choose 4 letters from the possible 5 unique letters (5 different ways) and multiple by the number of ways you can arrange them (4!). That's 5*24 = 120.
Number of words with duplicate a's: Use 2 a's plus 2 of the remaining 4 (4 choose 2). You can arrange those groups 4!/2! ways. That's 6*12 = 72. Add that to the 120 and you have 192 four letter words.
3 letter words:
Number of words without duplicates: Choose 3 letters from the possible 5 unique letters (10) and multiple by the number of ways you can arrange them (3!). That's 10*6=60
Number of words with duplicate a's: Use 2 a's plus 1 of the remaining 4 (4 choose 1). You can arrange those groups 3!/2! ways. That's 4*3 = 12. Add that to the 60 and you have 72 three letter words.
2 letter words:
Number of words without duplicates: Choose 2 letters from the possible 5 unique letters (10) and multiple by the number of ways you can arrange them (2!). That's 10*2=20
Number of words with duplicate a's: Just 1 (aa). Add that to the 20 and you have 21 two letter words.
Add them all together and you have your answer.