Liana has four cards which are numbered with the digits three, four, six, and nine. How many numbers greater than 9000 can she make by arranging the cards?
The easiest way of answering this problem is to list all the numbers that are greater than 9000. However, we will look at a second method that makes our life easier for more complex problems. As the number has to be greater than 9000, nine must be the first digit. The only way that the number can be greater than 9000 is if nine is in the thousands column. Putting any of the other three digits, three, four, or six in the hundreds column would create a number greater than 9000. Let’s start by putting the three in the hundreds column.
We are now left with the digits four and six which we can put in the tens and units column, respectively. This gives us the number 9346 which is greater than 9000. Whilst we could now move the digits three, four, and six around randomly, it is sensible to do so in a logical order. If we keep the three in the hundreds column, we can swap the six and the four. So the six is in the tens column and the four in the units. This gives us the number 9364. This is also greater than 9000. There are no other numbers we can create with nine in the thousands columns and three in the hundreds column. We know that nine has to stay in the thousands column.
Let’s now place the four in the hundreds column as shown. We can then place the three in the tens column and the six in the units column. This gives us the number 9436. Swapping the digits three and six gives us the number 9463. Both of these numbers are greater than 9000. We have now exhausted all the possibilities with three in the hundreds column and four in the hundreds column. We now need to try six in the hundreds column. Placing three in the tens and four in the units gives us 9634. Once again, we can swap the digits in the tens and units column. This gives us 9643.
These are all the possible combinations of the digits three, four, six, and nine that give a number greater than 9000. There are six different numbers, 9346, 9364, 9436, 9463, 9634, and finally 9643. As there were only six possible numbers here, this was a sensible way to approach the problem. However, if the problem was more complicated, there is a quicker method we could use.
Let’s firstly consider the number in the thousands column. We know that the number in the thousands column must be nine. Therefore, there is only one possible option. In the hundreds column, we had three choices. We could place the three, the four, or the six in this column. Once we’ve placed a number in the hundreds column, there are two possible choices in the tens column. Looking at the top two rows, once we’ve placed the three, we could place either four or six in the tens column. Once we’ve placed a number in the tens column, there is only one place left. Therefore, we only have one choice for the units column.
We can then calculate the total number of choices by multiplying these four numbers, the number of choices in the thousands column, the hundreds column, the tens column, and the units column. One multiplied by three multiplied by two multiplied by one is equal to six. Therefore, we have six choices in total. This confirms the six possible numbers that we found using the first method.