A hard question on combinations for P4–P5, with a step-by-step solution, diagrams and the answer. Counting arrangements and choices appears in the Hua Xia Cup from the P1 and P2 first rounds to the P5 finals.
The question
There are 5 digit cards:
0, 3, 4, 5, 8.
Each time, 3 of them make a three-digit number, each card used at most once, and 0 can't go in the hundreds place.
How many different three-digit even numbers can be made?
Show the hints (3)
💡Hint 1: Which digit decides whether a number is even? Deal with that place first.
💡Hint 2: Once the units digit is set, how many choices are left for the hundreds? Watch the 0: it may be used already, or still there.
💡Hint 3: Two cases: units 0 gives 12; units another even digit gives 18.
Worked solution
Have a go first, then read the solution step by step.
1.An even number ends in an even digit: 0, 4, 8.
2.Units digit 0: the hundreds can be any of the other 4, the tens any of the 3 left: 4 × 3 = 12.
3.Units digit 4, 8 (2 choices): the hundreds can't reuse that card or be 0, so 3 choices; the tens take any of the 3 left. 2 × 3 × 3 = 18.
4.In all: 12 + 18 = 30.
⚠️Common mistake: filling the hundreds first, then losing track of whether 0 is used. Fix the units first, and split cases for the 0.
Play more like this in Numsprite (unlimited questions, instant marking) →A free maths game: every level free from Easy to Expert (★ to ★★★★), unlimited questions, no download