Basic Counting Principles: Free Response

A gear setting is settled by two decisions, and each decision is a stage: first which chainring the chain sits on, then which sprocket it sits on at the back.

Every chainring can be paired with every sprocket, so whichever chainring is chosen, the back of the bike still offers all $11$ sprockets. That is exactly the condition the multiplication principle asks for, so the two stage counts multiply:

$3 \times 11 = 33.$

The bike offers $33$ gear settings. The $3$ counts the chainrings, the $11$ counts the sprockets available to each of them, and the product counts the pairs. A tree would draw $3$ branches, each splitting into the same $11$, giving $3$ equal groups of $11$ tips.

The bar changes what the back of the bike offers, but only for one of the chainrings, so the two stages no longer behave alike.

Sort the allowed settings by which chainring the chain sits on. A setting uses exactly one chainring, so those three groups share no member and every allowed setting sits in exactly one of them.

The two smaller chainrings are untouched by the manual and reach all $11$ sprockets. The largest chainring is barred from $2$ of them, so it reaches $11 - 2 = 9$.

The three group sizes add:

$11 + 11 + 9 = 31.$

There are $31$ gear settings. Notice why the product $3 \times 11$ is not available here. The second stage offers $11$ choices after two of the chainrings and $9$ after the third, so there is no single second-stage count to multiply by.

The multiplication principle is a statement about the number of choices, not about which choices they are. It applies to a sequence of stages when each stage offers the same number of choices whatever the earlier stages chose.

The bike in part B fails that test. The back of the bike offered $11$ sprockets after two of the chainrings and $9$ after the third, so there was no single second-stage count to multiply by, and the settings had to be gathered into groups and added instead.

Now take the second bike. Each of the $3$ chainrings is barred from $4$ of the $11$ sprockets, so each one reaches

$11 - 4 = 7$

sprockets. The four barred sprockets are different for each chainring, so the reachable lists really are different lists.

Draw the tree and you can see why that does not matter. Three branches leave the start, one for each chainring, and every one of them ends in seven tips, because every chainring reaches seven sprockets. The seven tips hanging off one branch are not the same seven as those hanging off another, but there are seven of them either way, so the tips still fall into $3$ equal groups of $7$, and counting equal groups is exactly what multiplication does:

$3 \times 7 = 21.$

The second bike offers $21$ gear settings, and its two stages are independent in the sense the principle means: which chainring the chain sits on does not change how many sprockets the next stage offers. Arranging items in order relies on the same distinction. Once a book has been placed on a shelf, which books remain depends on the one you placed, but how many remain does not, and it is the how many that the multiplication uses.

This is the part worth keeping: two stages can multiply even when the later stage's options change completely from one first-stage choice to the next. Only the count has to stay the same.

Build one phrase and watch what has to be decided. There are four rings, so there are four positions to fill, and each position is a stage.

Any of the $9$ bells can sound first. Once it has sounded it is used up and may not be used again in this phrase, so $8$ bells are available for the second position. That leaves $7$ for the third and $6$ for the fourth. Those counts hold whichever bells were used earlier, so the stages multiply:

$9 \times 8 \times 7 \times 6 = 3024.$

The choir can play $3024$ phrases. The counts fall by one at each step because there is one fewer unused bell to reach for, not because the bells themselves have changed.

Fill the position the rule pins down first. The opening ring must be the lowest bell, so there is exactly $1$ way to fill the first position.

That bell is now used up, and the remaining three positions are filled from the other $8$ bells with nothing else restricted: $8$ bells for the second position, then $7$, then $6$.

$1 \times 8 \times 7 \times 6 = 336.$

There are $336$ phrases that open on the lowest bell.

Filling the restricted position first is what keeps the later counts simple. Had the free positions been filled first from all nine bells, the number of bells left for the opening would depend on whether the lowest bell had already been used, and the stages would no longer offer the same number of choices whatever came before. Setting the lowest bell aside and then filling the free positions from the other eight is a different matter: that route is perfectly sound, and the alternate method takes it.

Read the count from left to right, one factor per position.

The first factor, $9$, counts the bells available for the opening ring, which is all of them. The second, $8$, counts the bells still unused when the second ring is chosen. The third, $7$, and the fourth, $6$, do the same for the third and fourth rings. Each factor drops by one because exactly one more bell has been used up by the time that position is filled.

So the last factor is $6$ rather than $9$ precisely because a phrase may not use a bell twice. Were the choir allowed to repeat bells, every position would keep all $9$ and the count would be a different and larger number:

$9 \times 9 \times 9 \times 9 = 9^4 = 6561.$

As for the order, the count follows a path through the four positions, and a path is a list of decisions taken in sequence. Ringing bell $2$, then bell $5$, then bell $7$, then bell $1$ is a different path from ringing bell $1$, then bell $7$, then bell $5$, then bell $2$, so the two reach different tips of the tree and are counted separately. They also sound different, which is the musical reason the choir cares. Every phrase corresponds to exactly one path and every path to exactly one phrase, so nothing is missed and nothing is counted twice.

Take the lanterns one at a time, in a fixed order, and treat each as a stage. The crew has plenty of gel in every color, so fitting a color to one lantern takes nothing away from the next: whatever the first three lanterns carry, the fourth still has all $7$ colors available.

The condition for multiplying is met, and the four stage counts are all the same:

$7 \times 7 \times 7 \times 7 = 2401.$

Four equal factors are exactly what a power abbreviates, so the same count is

$7^4 = 2401.$

There are $2401$ lighting plans. The base $7$ is the number of colors available at one lantern, and the exponent $4$ counts the lanterns.

The requirement changes what happens between the stages. A color fitted to one lantern is now spoken for, so the next lantern has one fewer color to choose from.

The first lantern may take any of the $7$ colors. The second may take any of the $6$ left, the third any of the $5$ still unused, and the fourth any of the remaining $4$. Each of those counts holds whatever the earlier lanterns took, so the stages still multiply:

$7 \times 6 \times 5 \times 4 = 840.$

There are $840$ plans in which the four lanterns carry different colors. The drop from $2401$ is the price of the requirement: every plan that repeated a color anywhere has been dropped.

Work both powers out before saying anything about either. One is four factors of $7$:

$7 \times 7 \times 7 \times 7 = 2401.$

The other is seven factors of $4$:

$4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 16{,}384.$

The two are built from the same pair of numbers and are nowhere near each other, which is worth registering on its own: swapping the base and the exponent is not a small slip.

Neither number is meaningless, though. Each counts a theater, and they are different theaters. The count $7^4 = 2401$ belongs to this one, where four lanterns each draw on the same $7$ colors: four stages, seven choices at each. The count $4^7 = 16{,}384$ belongs to a hall with seven lanterns fitted from a stock of only four colors, repeats allowed: seven stages, four choices at each.

What decides the roles is what each number counts. The base is the number of choices at a single stage; the exponent is the number of stages. In this hall the choices at one lantern are the $7$ colors and the stages are the $4$ lanterns, so the colors are the base and the lanterns are the exponent. Reading a repeated-choice situation in that order, choices first and stages second, fixes the power every time.

A member picks one walk, so what is wanted is the number of different walks printed anywhere in the two brochures.

Adding the two brochure lengths would be the addition principle, and that principle needs groups that share no member. These two lists do share members: $6$ walks are printed in both. So sort the walks into groups that do not overlap. Every walk falls into exactly one of these three.

Walks printed only in the morning brochure: that brochure lists $14$, of which $6$ are in the other one as well, leaving $14 - 6 = 8$.

Walks printed only in the wildflower brochure: $9 - 6 = 3$.

Walks printed in both brochures: $6$.

No walk belongs to two of those groups and none has been left out, so their sizes add:

$8 + 3 + 6 = 17.$

A member has $17$ walks to choose from.

Follow one shared walk through the volunteer's addition. A dawn chorus walk printed in both brochures is one of the $14$ counted from the morning brochure, and it is also one of the $9$ counted from the wildflower brochure. The addition counts that single walk twice, although a member can sign up for it only once.

The same happens to each of the $6$ walks printed in both, and to none of the others. So the volunteer's total carries exactly $6$ extra copies:

$23 - 6 = 17.$

The volunteer's method is not wrong in general; it is wrong for these two lists. Adding brochure lengths counts printings, and a walk printed twice is still one walk. What a member wants to know is how many different walks are on offer.

The addition principle says that when exactly one item is chosen and the groups share no member, the number of choices is the sum of the group sizes. The condition is the part doing the work: no item may belong to two groups, so that adding the sizes counts each item exactly once.

The two brochures are not such groups, because $6$ walks belong to both. The center could relist the same walks in three groups that are: walks in the morning brochure only, walks in the wildflower brochure only, and walks in both. Every walk lands in exactly one of the three and nothing is left out, so the principle applies to that listing directly and the three sizes add:

$8 + 3 + 6 = 17.$

Could the two brochure lengths ever have been added? Yes, but only in a case the center does not have. If no walk were printed in both, the two lists would share no member, the condition would hold, and $14 + 9$ would be the right count. Each shared walk costs the sum exactly one extra copy, so the two lists give the right total precisely when they share nothing, and the gap here is the number of shared walks and nothing else.

An assignment names two things, a stretch and a day, so building one is a sequence of two stages. Every stretch can be surveyed on every day, so whichever stretch is named, the assignment still has all $7$ days available, and the stage counts multiply:

$9 \times 7 = 63.$

The box must hold $63$ slips. A grid makes the same count visible: $9$ rows of stretches by $7$ columns of days has $63$ cells, each cell is one assignment, and the box holds one slip per cell.

Adding a third item to every assignment adds a third stage, and the recording method does not depend on what the first two stages named: all $3$ methods are available for any stretch on any day. So a third factor joins the product:

$9 \times 7 \times 3 = 189.$

The box would hold $189$ slips.

The step from one season to the next is worth seeing on its own. Each slip from the first box becomes $3$ slips, one for each recording method, so the total is multiplied by $3$ rather than increased by $3$.

Filling the box the second way makes a differently shaped tree. The first way starts with $9$ stretch branches and splits each of them into $7$ days. The second starts with $7$ day branches and splits each of them into $9$ stretches.

The totals are

$9 \times 7 = 63, \qquad 7 \times 9 = 63,$

the same number, because the order of two factors does not change a product.

More than the totals agree. In the first tree a slip is reached by naming a stretch and then a day; in the second, by naming a day and then a stretch. Either way the slip that comes out says the same thing: this stretch on this day. Each of the possible assignments is reached by exactly one path in each tree, so the two boxes hold the very same slips, sorted differently.

Which stage you count first is therefore a matter of how you organize the work, not of how many possibilities there are. That is a useful thing to know when one stage is easier to think about than the other: you may start from whichever you like.