Digits Arrangement

This item is taken from IGCSE Additional Mathematics (0606) Paper 22 of October/November 2010.
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To deal with this item, let us summarize the details of the problem:
   (a) Syllabus area: Permutation and Combination 
   (b) Formula/Concept needed: 
            > Permutation or Combination
            > Fundamental Principle of Counting
            > Even numbers
            > No Repetition of Digits

Before we answer the items, let us encircle the important clues/phrases.

It means that

The next phrase is

It means that

Now, let us answer (i)

Note that the number has four (4) digits

Since there are no restrictions (except for the no repetition), we are considering all possible numbers that could be formed using the given digits. The first digit then can have

After using one on the first digit, the second digit will have

The third digit will then have

And the last digit will have

Multiplying the options for each digit,

Hence, there are 840 (4-digit) numbers that could be formed from the given options.

Another way of solving this item is to use Permutation (nPr). This is possible since order is important. That is

For (ii)

To make sure that the number is less than 4000, we will start with the first digit. A number is less than 4000 if the first digit is a 1, 2, or 3. However, in the given options, there are only two (1 or 3).

The remaining could be any digit,

Multiplying the options,

Hence, there are 240 (4-digit) numbers that could be formed that are less than 4000.

For (iii)

Again, to make sure that it is less than 4000, we will start with the first digit.

Further, to make sure that it is even, the last digit must contain an even digit. In the options given, only 4 and 8 are even.

The second and third may now have any of the remaining digits.

Multiplying the options,

Hence, there are 80 (4-digit) numbers that could be formed that are even and less than 4000.
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