[Math] How many different 3-digit numbers can be formed if the number can consist of the digits from 0 to 7, cannot start with 0 and must be even

permutations

Original question:

A 3-digit number is made up using the digits 0, 1, 2, 3, 4, 5, 6 and 7 at most once each. The number cannot start with 0. How many different numbers can be formed if the number must be even?

Approach:

Working backwards, there are four possibilities for the final digit since the final digit must, for the number to end up even, be 0, 2, 4 or 6. There are seven possibilities for the penultimate digit since there are eight digits that can be used but one of them has been used up on the final digit (because no digit can be repeated within a number). There are five possibilities for the first digit because two digits have already been used up and zero is invalidated due to the question's stipulation that the number cannot start with zero (which would have been pretty self-explanatory anyway).

4 x 7 x 5 = 140 different numbers which can be formed.

However, the textbook says that 150 different numbers can be formed. Where have I gone wrong?

Best Answer

Suppose for example that you choose

last digit $4$, second-last digit $0$.

Then there are $6$ possible choices for the first digit, not $5$ as you claimed.

To solve the problem it's probably best to split into two cases:

  • if the last digit is zero then there are $7$ possibilities for the first digit and $6$ for the second, total $42$;
  • if the last digit is not zero then there are $3$ choices for the last digit, $6$ for the first and $6$ for the secong, total $108$.

Final total, $150$.

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