Problem
Addition and multiplication counting rules
A box holds 4 blue balls and 3 yellow balls. Part (1) takes one ball that is blue or yellow. Part (2) takes two balls, one blue and one yellow. Count the ways for each part.
Reasoning · Grade 5-2 Counting Cases
Your answer
Watch out The parts invite the same method, but 'or' adds while 'and' multiplies.
How to solve
Strategy Make a Systematic List — The numbers here are small enough to list every single case, and listing them is what makes the difference between the two parts visible instead of just remembered. I give the blue balls names B1, B2, B3, B4 and the yellow balls Y1, Y2, Y3, then write out the possible results for each part. For part (2) I organise the list as a tree diagram, one branch for each blue ball splitting into three yellow branches, and also as a 4-row by 3-column table, so I can count the same 12 pairs two different ways. Once the lists are on paper the two shortcut rules — add when the events cannot both happen, multiply when they happen together — come out of the counting rather than being taken on trust.
1STEP 1
Give the balls names so they can be counted
Give each ball a name.
blue: B₁, B₂, B₃, B₄ yellow: Y₁, Y₂, Y₃
Naming identical-looking objects is the standard trick that turns a fuzzy 'how many ways' question into a list you can actually write down and count.
7.SP.C.8Make A Systematic List2STEP 2
Part (1): list every way one ball can come out
Listing one-ball results gives 7.
B₁, B₂, B₃, B₄, Y₁, Y₂, Y₃ → 7
When you can write out every possibility on one line, counting the line is a complete answer that needs no rule at all.
7.SP.C.8Make A Systematic List3STEP 3
Part (1) again with the addition rule
It is an 'or', so 4 + 3 works too.
4 + 3 = 7
Adding is exactly what joining two separate piles of possibilities does, and a child can see the join is safe because no ball appears in both piles.
1.OA.A.1Organize Information In More Ways4STEP 4
Part (2): grow a tree of blue-and-yellow pairs
The tree gives three yellows per blue.
B₁ → Y₁, Y₂, Y₃ B₂ → Y₁, Y₂, Y₃ B₃ → Y₁, Y₂, Y₃ B₄ → Y₁, Y₂, Y₃
A tree diagram makes 'and' visible: you cannot stop at the first level, you have to walk to the end of a branch, and every path from root to tip is one complete case.
7.SP.C.8Draw A Diagram5STEP 5
Count the branch tips with a multiplication
The tips number 4 × 3 = 12.
4 × 3 = 12
Equal groups are what multiplication is for, and the 4 by 3 rectangle of pairs is the same array picture used to learn the times tables.
3.OA.A.1Organize Information In More Ways6STEP 6
Say clearly why one part adds and the other multiplies
So one part adds and the other multiplies.
or → 4 + 3 = 7 and → 4 × 3 = 12
Deciding whether the two events can happen at the same time is the whole decision; once that is settled the arithmetic is a single addition or a single multiplication.
7.SP.C.8Make A Systematic ListOne part adds because the cases cannot both happen, and the other multiplies because the two choices are made together.
Why?
Drawing one ball gives either a blue or a yellow and never both, so the two counts cover everything without overlapping.
🧱Disjoint cases addSplit the possibilities so no case overlaps another and none is left out, and the case counts simply add.
Why?
Drawing a blue and a yellow means both choices are made, and neither limits the other, so the counts multiply.
🧱Independent choices multiplyIf a first choice can go a ways and each of those leaves b ways for a second, the pair can be made a times b ways.
Answer
7, 12 ways
4 + 3 = 7, 4 × 3 = 12
Part (1) had to come out as 7, because taking one ball out of a box of 7 balls can only end 7 ways, and every one of those balls is either blue or yellow — so the answer counts all the balls exactly once, which is a good sign. Part (2) must be bigger than part (1), since each of the 12 pairs already contains one blue ball and one yellow ball, and 12 is indeed more than 7. It also has to be smaller than the 21 pairs you would get from any two of the 7 balls, because same-colour pairs are not allowed; 12 sits sensibly below 21. Checking the pairs a second way, as unordered two-ball handfuls with one of each colour, again gives 12. Both answers are whole numbers of cases, which is the only kind of answer a counting question can have.
Takeaway
Look for the little word: 'or' means the cases sit side by side so you add them, and 'and' means every choice pairs with every choice so you multiply!
- Give the balls names so they can be counted
- Part (1): list every way one ball can come out
- Part (1) again with the addition rule
- Part (2): grow a tree of blue-and-yellow pairs
- Count the branch tips with a multiplication
- Say clearly why one part adds and the other multiplies