3
$\begingroup$

How many four-digit odd numbers, all of digits different, can be formed from the digits 0 to 9, if there must be a 5 in the number?

I know that there are 4 different cases where 5 is in the number:

5 _ _ _

_ 5 _ _

_ _ 5 _

_ _ _ 5

The first digit cannot be 0, since it has to be a 4-digit number.

$\endgroup$
  • $\begingroup$ Is it possible for 5 to appear more than once in the number? $\endgroup$ – John Molokach Jun 12 '15 at 20:07
  • $\begingroup$ @JohnMolokach "all digits different" $\endgroup$ – Joffan Jun 12 '15 at 20:10
8
$\begingroup$

I would fill in the last number first and consider two cases: the last number is a $5$, and the last number is not a $5$.

Case 1: Last number is five.

This is the easy case, the remaining digits can be any distinct digits that are not $5$ (and do not begin with zero). This gives $8$ choices for the first number (not 5 and not 0), $8$ choices for the second number (not the first number, and not $5$) and $7$ choices for the third number (not the first two, and not $5$--the last). This gives $8\cdot 8\cdot 7=448 options.

Case 2: The last number is not five.

First, we will make sure the number is odd. To do that, the last digit must be either $1, 3, 5, 7,$ or $9$. Since it is not $5$, we have $4$ choices for the last digit. Now, let us fill in the first digit (since it cannot be zero). There are again $8$ choices for the first digit (not the last digit and not zero). There are again, $8$ choices for the second number and $7$ for the third (for exactly the same reasons as mentioned before). We now have $4\cdot 8\cdot 8\cdot 7=1792$ numbers. However some of these numbers didn't use a $5$. How many didn't use $5$? Let's count that and subtract. There are again $4$ choices for the last number (since $5$ wasn't an option there). Now there are only $7$ choices for the first number: not $5$, not $0$, and not the last digit. Similarly, there are $7$ choices for the second digit and $6$ choices for the third digit. Thus, there are $7\cdot 7\cdot 6\cdot 4=1176$ numbers that don't use $5$ at all. Thus we have $1792-1162=616$ good numbers.

Combining case 1 and case 2 gives us $448+616=1064$ four-digit, odd numbers with distinct digits and a $5$ occurring exactly once.

$\endgroup$
5
$\begingroup$

There are special conditions attached to the $\color{red}{\text{first}}$ and $\color{blue}{\text{last}}$ characters, so group into 3 cases:

  • $5$ first: then for the remaining places there are (choosing the last digit first) $\color{blue}{4}\times 8\times 7 = 224$ options
  • $5$ second/third: then for the remaining places there are $\color{blue}{4}\times \color{red}{7}\times 7 = 196$ options for each, so total of $2 \times 196 =392$ for this case.
  • $5$ last: then for the remaining places there are $\color{red}{8}\times 8\times 7 = 448$ options

Total options then are: $224+392+448=1064$.

$\endgroup$
1
$\begingroup$

Another way to do this is to think of all the cases generally, and subtract all cases without any $5$s, because your question is essentially asking, "How many 4-digit odd numbers, all digits different, can we form where one of the digits is $5$?"

So, the number of 4-digit odd numbers in general is $8\cdot8\cdot7\cdot5$. Explanation: $5$ possible numbers in the last slot to make it odd, $8$ in the first slot because it can't be what was in the last slot and it can't be $0$, $8$ in the second slot (neither first nor last, but you can use $0$), and then $7$ in the third slot.

The number of 4-digit odd numbers without any $5$s is $7*7*5*4$. Explanation: $4$ possible numbers in the last slot to make it odd, sans $5$. Follow same arguments as above.

Final answer: $8\cdot8\cdot7\cdot5 - 7\cdot7\cdot5\cdot4=1064$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.