Combinatorics: What to Do with "Forbidden Choices"
How many four-digit numbers are there where the first digit is even, the second is odd, the third is a prime number, the fourth (units digit) is divisible by 3, and the digit 2 can be used at most once?
Now, define the __forbidden combinations__: the phrasing "the digit 2 can be used at most once" means that the __good combinations__ include those numbers where the digit 2 either does not appear at all or appears just once.
Therefore, the forbidden combinations include the remainder: those scenarios in which the digit 2 appears more than once. Since 2 can _only appear in the thousands and tens digits_, the forbidden combinations include those numbers where both of those digits are 2.
We have to count the number of forbidden cases in which 2 appears twice. To do that, we need to assume that 2 _does_ appear twice and check the other places:
1. The first digit must be 2: __1 option__ (2).
2. The second digit can be any odd digit: __5 options__ (1, 3, 5, 7, 9).
3. The third digit must be 2: __1 option__ (2).
4. The fourth digit can be any multiple of 3: __4 options__ (0, 3, 6, 9).
That's a total of 20 forbidden options in which 2 appears twice:
>$$\mathop{\fbox{${1}$}}\limits^{{}{}\hspace{0.33em}{\text{I}}}
\times \mathop{\fbox{${5}$}}\limits^{{}{}\hspace{0.33em}
{\text{II}}} \times \mathop{\fbox{${1}$}}\limits^{{}{}\hspace{0.33em}{\text{III}}}
\times \mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}{\text{IV}}} = 20$$.
Finally, we need to subtract the forbidden choices from the total choices to find the good options—numbers in which the number 2 appears only once or does not appear at all:
>$$\text{Good combinations} = 320-20=300 \text{ options}$$.
Incorrect.
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Note that you have to consider the fact that the number 2 cannot appear more than once.
Incorrect.
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Incorrect.
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Make sure that for each __SeBox__ you assigned the correct number of items that meet the terms.
Think again—how many numbers between 0 and 9 are divisible by 3?
Incorrect.
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Correct.
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Find the total number of combinations, disregarding the restrictions on the number 2, by breaking down the problem set by step using __SeBoxes__.
The options for the digits are as follows:
1. The first digit must be even but not zero: __4 options__ (2, 4, 6, 8).
2. The second digit must be odd: __5 options__ (1, 3, 5, 7, 9).
3. The third digit must be prime: __4 options__ (2, 3, 5, 7).
4. The fourth digit must be divisible by 3: __4 options__ (0, 3, 6, 9). _Remember that 0 is divisible by all positive integers_.
That gives us a total of 320 options:
>$$\mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}{\text{I}}}
\times \mathop{\fbox{${5}$}}\limits^{{}{}\hspace{0.33em}
{\text{II}}} \times \mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}{\text{III}}}
\times \mathop{\fbox{${4}$}}\limits^{{}{}\hspace{0.33em}{\text{IV}}}=320 $$.
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