It's a basic question, but I don't know why I am getting confused.
Determine the total number of $4$-digit numbers which can be obtained using the digits $1, 2, 3, 4, 5$. Also find how many of them are divisible by $4$.
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How many choices do you have for the first digit? How many for the second? – Tad Oct 18 '15 at 02:46
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I have written the question as it is. There's no restriction on digits' choice. – Paras Sidhu Oct 18 '15 at 02:50
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Sorry, that was intended to be a hint. – Tad Oct 18 '15 at 02:50
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The answer to your question depends on whether a digit may be repeated. For instance, is the number $1244$ permitted? – N. F. Taussig Oct 18 '15 at 08:58
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@N.F.Taussig It's a question of my math exam of last year. Since there is no clue of repetition, we may assume repetition is allowed. Can you help now? – Paras Sidhu Oct 18 '15 at 09:02
2 Answers
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Case 1: digits cannot be repeated
For the first question, there are five choices for each of the four digits.
For the second question, there are five choices for the thousands digit and five choices for the hundreds digit. For a number to be a multiple of $4$, the last two digits must be a multiple of $4$. How many such multiples of $4$ are there between $11$ and $55$ inclusive?
Case 2: digits cannot be repeated
For the first question, there are five choices for the thousands digit, four choices for the hundreds digit, three choices for the tens digit, and two choices for the units digit.
For the second question, how many multiples of $4$ that lie between $11$ and $55$ inclusive do not contain a repeated digit? Since each of these choices require the use of two distinct digits, there are three choices for the thousands digit and two choices for the hundreds digit.
N. F. Taussig
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1For the first question, the answer is $5 \cdot 5 \cdot 5 \cdot 5 = 5^4$. What do you get for the second question? – N. F. Taussig Oct 18 '15 at 09:45
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1oh yes. Thanks! For second one there are 11 numbers multiple of 4 between 12 and 54. Although the answer of second question will be same, we should have looked in 11 and 55 inclusive given that repetition is allowed, right? – Paras Sidhu Oct 18 '15 at 09:52
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You are quite right. The reason I looked between $12$ and $54$ is that a multiple of $4$ must be even. Since there are $5$ choices for the thousands digit, five choices for the hundreds digit, and $11$ choices for the final two digits, we obtain $5 \cdot 5 \cdot 11$ for the second question. – N. F. Taussig Oct 18 '15 at 09:57
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How many ways can you pick a digit to not use?
For each of these, how many ways can you arrange the remaining four digits?
A number is divisible by $4$ if and only if the last two digits (on the right) are divisible by 4.
There are twenty possible two-digit pairs, from $12$ to $54$. How many of them are divisible by $4$?
DJohnM
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Your answer to the first question assumes digits cannot be repeated. Your answer to the second question seems to permit the repetition of digits. – N. F. Taussig Oct 18 '15 at 09:02