Problem
Draw a table to solve
46 students must all sit at 4-seat and 6-seat tables with no empty seat. Each table count is a whole number, zero or more. Count the different ways the table numbers can be chosen.
Reasoning · Grade 2-2 Problem Solving with Tables
Your answer
Watch out Using zero six-seat tables looks allowed, but 46 does not divide by 4, so a seat would sit empty.
How to solve
Strategy Make a Systematic List — We need every whole-number combination of 4-seat and 6-seat tables that seats exactly 46 people with no empties. Listing the choices in order by the number of 6-seat tables, from fewest to most, lets us check each case once and count without missing or repeating any.
1STEP 1
Set up the seat total
With no empty seat the seats must total exactly 46.
4 × (4-seat tables) + 6 × (6-seat tables) = 46
Each table fills with a fixed group size, so we just need groups of 4 and groups of 6 that combine to 46.
3.OA.A.3Make A Systematic List2STEP 2
Try each number of 6-seat tables in order
Step the six-seat count up from 0 and test each remainder against 4.
46-6=40=4×10; 46-18=28=4×7; 46-30=16=4×4; 46-42=4=4×1
Stepping the 6-seat tables up one at a time means we never skip a possibility and never count one twice.
3.OA.A.4Look For A PatternStepping the number of 6-seat tables up one at a time reaches every possibility once and none of them twice.
Why?
A seating plan has one definite number of 6-seat tables, so it falls into exactly one row of the sweep.
🧱Disjoint cases addSplit the possibilities so no case overlaps another and none is left out, and the case counts simply add.
Why?
For each choice, what is left of the 46 seats must break into whole groups of 4, and any leftover kills that row at once.
🧱Division with remainderAny number splits into a whole count of equal groups plus a leftover, and the leftover is always smaller than the group size.
3STEP 3
Collect the working combinations
The ones that work are (10,1), (7,3), (4,5), (1,7) — four ways.
(10,1), (7,3), (4,5), (1,7) → 4 ways
Counting the rows of our organized list gives the number of ways directly.
2.OA.A.1Make A Systematic ListAnswer
4 ways
(4-seat, 6-seat) = (10,1), (7,3), (4,5), (1,7)
Check each combination seats exactly 46: 10x4 + 1x6 = 40+6 = 46; 7x4 + 3x6 = 28+18 = 46; 4x4 + 5x6 = 16+30 = 46; 1x4 + 7x6 = 4+42 = 46. All four work, and the cases in between (0, 2, 4, 6 six-seat tables) leave a remainder that is not a multiple of 4, so they are correctly rejected. Exactly 4 ways.
Takeaway
Listing the 6-seat tables in order, none-then-one-then-two, lets you find every seating with no empty chairs - just Grade 3 grouping you already know!
- Set up the seat total
- Try each number of 6-seat tables in order
- Collect the working combinations