Mathematical reference
- Four objects, choose two
- C(4,2) = 6 unordered pairs, each with one distinct four-step route
- Partition by the fifth object
- C(5,2) = C(4,1)+C(4,2) = 4+6 = 10
- Complementary choices
- Choosing 2 of 5 fixes the 3 left out: C(5,2)=C(5,3)
Lesson transcript
This is the text of the narrated lesson. References to turning, highlighting or changing an object describe the linked interactive lesson; its demonstrations are not embedded in this reading page.
Give each step an object
Begin with four distinct objects, named A, B, C, and D. Keep them in that fixed order. We will make one decision for each object: take it or leave it. The highlighted route records those decisions. Moving down-right means take the next object; moving down-left means leave it. Four decisions therefore make a route with four steps.
Follow this route and read the group beside it. Right, left, right, left means take A, leave B, take C, leave D. The selected group is A and C. The letters inside braces name the members of the group. They do not describe the order in which you might physically pick those objects up.
Try the route with two left steps followed by two right steps. We leave A and B, then take C and D. The group changes to C and D. The destination stays in row four, position two, because there are still four decisions and exactly two right steps. Different groups can reach the same counting cell.
Now put the two right steps first. We take A and B, then leave C and D. The group is A and B. Again the route ends at the six. That entry counts how many different two-object groups are possible; it does not say there are six objects inside this particular group.
See every pair exactly once
Watch two of the possible pairs in succession: A with B, then A with C. The readout changes with the route. Both pairs contain A, but the second selected object differs. Keep the four object positions fixed while you read the steps. Changing which object occupies a position would change the meaning of the route.
The next two pairs are A with D, then B with C. Follow the right steps to locate their members. In the first route, the right steps are at the beginning and end. In the second, they are the two middle steps. Both routes have two rights, and both end at the same six.
Complete the list with B and D, then C and D. We have now visited all six pairs from these four objects. Each pair has its own route. The selected cell counts the whole family, while the blue line shows one member of that family at a time. Do not confuse the highlighted route with the complete count.
Let all six routes pass once more. You can read the changing pair and match it to the two right steps. There are six possibilities because the pair of selected positions can vary in six ways. We are not multiplying six unrelated examples: we are enumerating one collection, with no pair repeated and none omitted.
Translate a group into its route
Extend the fixed list to five objects, A through E. Choose A and D. To recover its route, scan the entire list in order: take A, leave B, leave C, take D, leave E. Each membership decision determines one step. The group fixes the route completely, even though the braces themselves do not impose a picking order.
Read a different route in the other direction: left, right, left, left, right. Its right steps occur at the positions of B and E, so the selected group is B and E. The route determines the group completely as well. These two translations undo one another; they do not create extra choices or lose information.
Now select three objects: A, C, and E. The route alternates right and left, beginning and ending with right. It has five steps and three rights, so it reaches row five, position three. The same correspondence works for triples. Only the required count of right steps has changed.
Compare with the triple A, B, and C. Its first three steps go right, followed by two lefts. Both triples reach the ten in position three. The endpoint records the size of the selected group, while the full route records exactly which objects were selected. We need both descriptions to distinguish a family from one of its members.
Partition by the final decision
Return to the pair count in row five, position two. In Parents, its contributions are four and six, giving ten. Think about the fifth object, E. Every pair either includes E or leaves it out. Those are different possibilities, and together they account for every pair. The routes will make their two incoming families visible.
Watch the four pairs that include E. The other member is A, B, C, or D. Every one of these routes ends with a right step, because the last decision takes E. Before that final step, the route has selected one object from the first four. That is why this family comes through the parent four.
Now watch the six pairs that leave E out. Both members come from A, B, C, and D. Each route ends with a left step, leaving the selected pair unchanged on the final decision. Before that step, two objects have already been selected from the first four. These routes come through the parent six.
Return to the two parents. Four routes arrive by the final right edge, and six by the final left edge. No route can have both final steps, and every route has one of them. Adding four and six therefore counts all ten routes and all ten pairs. This is the parent rule from the first lesson, now read as a partition of choices.
Choosing a group also chooses what stays out
Select the pair A and C from five objects. Its route is right, left, right, left, left. The objects left out are B, D, and E. Once the selected pair is fixed, that complementary triple is fixed too. There is no additional decision to make about which objects remain outside the group.
Select that complementary triple instead. The right and left decisions exchange roles: left, right, left, right, right. We now take B, D, and E and leave A and C. The destination moves from position two to position three. Every pair corresponds to one complementary triple, with a route obtained by swapping the two step directions.
Switch to Choose and look at the two highlighted tens in row five. One counts pairs and the other counts triples. Their equal values come from the complement correspondence we just followed. The mirror symmetry is useful to see, but the matching groups explain why the counts agree.
Try another complementary pair of groups. We select D and E here, leaving A, B, and C outside. The highlighted route takes only the last two objects. As an experiment, name the complementary triple before changing the view yourself. Swapping take and leave will reach the reflected position in the same row.
One empty group and one complete group
Return to four objects and leave every one of them out. The route has four left steps and ends at the outside one. The braces are empty because the selected group has no members. There is exactly one empty group, so this count is one rather than zero. Zero members and zero possible groups are different statements.
Take all four objects instead. The route has four right steps and reaches the other outside one. There is exactly one complete group containing A, B, C, and D. These two routes explain the boundary ones from the first lesson: take nothing, or take everything, each in one way.
Move one position inward and select just one object. The four routes take A, B, C, or D in turn. Each has a single right step, placed at a different point in the fixed list. The destination is four because there are four distinct one-object groups. Here the count is especially easy to check directly.
Return to the whole triangle in Choose. The boundary one and its reflected one match, just as the two inner fours match. A group and its complement always use the same original set of objects. Comparing different original sets would ask a different counting question, even if two displayed values happened to agree.
Separate one example from the whole count
Extend the list to six objects, A through F, and select A and F. The route takes the first and last objects and leaves the others out. It reaches fifteen in row six, position two. This is one example among fifteen possible pairs. We can display one route clearly without pretending that one visible route is the entire family.
Change the selected pair to C and D. The two right steps move to the middle of the six-step route. The endpoint remains fifteen because the group size is still two. This change makes a different member of the same family. It does not add to the total fifteen or alter the triangle’s coefficients.
Select three of the six objects: A, C, and F. The route now has three right steps and ends at twenty in row six, position three. Twenty is the count of all such triples. The new destination reflects a changed group size, while the particular positions of the three rights identify this one triple.
Finish with Choose at the same twenty. The triangle can be read as counts of selected groups, or as counts of routes with a specified number of right steps. These counts agree because each group determines exactly one route and each route determines exactly one group. The parent rule separates the families by their last decision. Keep that connection as we move on to sums along the diagonals.