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How many four-digit positive integers are there that have the digit 1?

attempt: There are $9000$ numbers that contains $4$ digits. Consider the case where we have $abcd$ where $abcd$ represent the digits. Then we would have $9$ choices for $a$ and $10$ choices for $b,c$ and assume last digit has $d = 1$ , so we would have $9(10)(10)(1)$ choices . Similarly if we assume at least $c = 1$, then we have $9$ choices for $a$ and $10$ choices for $b,d$ , so we have $9(10)(1)(10)$ choices, similarly when we consider the case if $a = 1$ and $b = 1$. Then we would to get the total four-digit positive integers such that contain the digit 1 to be $3700$. Can someone please verify this makes sense? Thank you!

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    $\begingroup$ I believe that you are over-counting. Assume $d=1$. Then you have $(9)(10)(10)(1)$ choices. One of these choices is the number $1111$. Now assume $c=1$. You state that we have $(9)(10)(1)(10)$ choices. But one of these choices is the number $1111$. So we've counted the number $1111$ twice. $\endgroup$
    – Tim Thayer
    Feb 16, 2017 at 6:19

2 Answers 2

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Instead, consider the following question: How many four digit numbers don't have the number $1$?

To do this, if the number is $abcd$, then $a$ has $8$ choices, and the rest have $9$ choices, since $a \neq 0,1$. Then, we get $8 \times 9^3 = 5832$. Subtracting this from the number of $4$ digit numbers, which is $9000$, gives the answer $3168$.

So where is your mistake? Well, I can't say, because when I followed your calculation I did not end up with $3700$. So elaborate and show your calculations, and I'll edit the answer if need be.

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    $\begingroup$ I agree with 3168. $\endgroup$
    – Tim Thayer
    Feb 16, 2017 at 6:20
  • $\begingroup$ @TimThayer Thank you, I have verified the same. Having said that, are you able to understand how the answer $3700$ came up? It is important to point out the mistake in the procedure as well, rather than just give the final answer. $\endgroup$ Feb 16, 2017 at 6:21
  • $\begingroup$ I am still having trouble understanding why you are only considering the case for $a$ to be from ${\{2,...,9}\}$. And how that counts the numbers that don't contain a one. What about the rest of digits? $\endgroup$
    – Mahidevran
    Feb 16, 2017 at 6:23
  • $\begingroup$ @Mahidevran See, the number must be four digits, so $a$ can't be zero, otherwise it'll get reduced to $3$ or lesser digits. Also, $a$ cannot be one, because we are considering the cases when none of the digits are one. However, for $b,c$ and $d$, these can happily be zero, right? (For example, $4000$, where $b,c,d = 0$). However, by what we want, these can't be $1$, so $b,c,d \in \{ 0,2,3,4,5,6,7,8,9\}$, which gives $9$ choices to each of $b,c,d$, and $8$ choices to $a$. Then, because these digits can be chosen independently, we can multiply them to get the total non-probability. $\endgroup$ Feb 16, 2017 at 6:26
  • $\begingroup$ @астонвіллаолофмэллбэрг Using the (incorrect) logic provided, when $d=1$, there are $(9)(10)(10)(1)=900$ choices. When $c=1$, there are $(9)(10)(1)(10)=900$ choices. When $b=1$, there are $(9)(1)(10)(10)=900$ choices. When $a=1$, there are $(1)(10)(10)(10)=1000$ choices. The total number of choices is 3700. $\endgroup$
    – Tim Thayer
    Feb 16, 2017 at 6:28
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As @tim thayer said You have repeated some numbers like 1111 It will appear in the four possible places of 1 So your answer is not pricise

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