QuanticaPrealgebraOpen in the courseLast lesson we counted by adding and subtracting whole piles of items. Now we count outcomes that get built up in steps, like an outfit picked one piece at a time, where each step has its own set of options.
Problem
You have \(5\) shirts and \(4\) pairs of pants. How many different shirt-and-pants outfits can you make?
Show a hint
- Pick just one shirt for a moment and hold it still. With that single shirt in your hand, how many different outfits can you build by trying each of the \(4\) pairs of pants with it?
- One shirt gives \(4\) outfits. Now you have \(5\) shirts, and every shirt gets its own group of \(4\) outfits, with no outfit ever shared between two shirts. So you are stacking up \(5\) groups that each hold \(4\) outfits.
- You have \(5\) equal groups of \(4\) outfits, so add \(4+4+4+4+4\), which is the same as the product \(5 \times 4\).
Show the full solution
Hold one shirt still. It pairs with each of the \(4\) pairs of pants, so that single shirt gives \(4\) outfits. Every one of the \(5\) shirts gives its own \(4\), and none of those repeat, so the total is \(5 \times 4 = \boxed{20}\). Equal-sized groups get multiplied, which beats adding \(4+4+4+4+4\) by hand.
Problem
A game lets you pick one of \(2\) body types, \(3\) hair styles, \(4\) outfits, and \(6\) weapons. How many different heroes can you build?
Show a hint
- Building a hero is four separate decisions in a row, one for each menu, so set up four slots and write the number of choices that fit in each one.
- When a job is made of stages done one after another, and the choices at each stage do not change the choices at the next, you multiply the slot numbers together instead of adding them.
- Your four slots hold \(2\), \(3\), \(4\), and \(6\), so multiply \(2 \times 3 \times 4 \times 6\) to get the total.
Show the full solution
Four menus means four slots, holding \(2\), \(3\), \(4\), and \(6\) choices. Multiply straight across. \(2 \times 3 = 6\), then \(6 \times 4 = 24\), then \(24 \times 6 = 144\), so you can build \(\boxed{144}\) heroes. Picking a body type does not remove any hair or outfit options, so every stage keeps its full count and the slots multiply.
Problem
A lock has \(3\) dials, each showing one of \(6\) colored symbols. Repeats are allowed. How many \(3\)-symbol codes are possible?
Show a hint
- Picture the lock as \(3\) slots side by side. For each slot, ask how many symbols could go there. Since a symbol can repeat, choosing it for one dial does not stop you from using it again on the next.
- Use the multiplication principle. Count the choices for the first dial, then the second, then the third, and multiply those counts together to get the total number of codes.
- Each dial has \(6\) choices, and the choices stay the same because repeats are allowed. So the total is \(6 \times 6 \times 6\). Work out that product.
Show the full solution
Each of the \(3\) dials can show any of the \(6\) symbols, so multiply \(6 \times 6 \times 6 = 6^3 = 216\). The lock has \(\boxed{216}\) codes. Repeats being allowed is what keeps the count at \(6\) every time, since using a symbol on one dial does not use it up.
Problem
A license plate is 2 letters then 2 digits. Repeats allowed. How many different plates are possible?
Show a hint
- Picture the plate as four empty slots in a row, two for letters and two for digits. Each slot gets filled on its own, so think about how many choices live in each one.
- Use the multiplication principle. Count the choices for each slot separately, then multiply all four counts together. Because repeats are allowed, a slot does not lose any choices from the slots before it.
- A letter slot has \(26\) choices and a digit slot has \(10\) choices, so the count is \(26 \times 26 \times 10 \times 10\). Multiply these to finish.
Show the full solution
Four slots, letter, letter, digit, digit. A letter slot has \(26\) choices and a digit slot has \(10\), and repeats are allowed so no slot loses anything to the ones before it. The two letters give \(26 \times 26 = 676\), the two digits give \(10 \times 10 = 100\), and \(676 \times 100 = 67600\). There are \(\boxed{67600}\) plates.
Problem
How many different orders can \(6\) different books be arranged on a shelf?
Show a hint
- Picture the shelf as \(6\) empty slots in a row, one for each spot from left to right. Filling the slots one at a time is far easier than trying to list every arrangement by hand.
- Use the slot-and-box habit. Count the choices for the first slot, then the second, and so on, and multiply all the counts together. The key is that once a book is placed it is gone, so each slot has one fewer choice than the slot before it.
- The first slot has \(6\) books to choose from, the next has \(5\) left, then \(4\), then \(3\), then \(2\), and the last slot has just \(1\) book remaining, so multiply \(6 \times 5 \times 4 \times 3 \times 2 \times 1\).
Show the full solution
Fill \(6\) shelf slots left to right. The first slot has \(6\) books to choose from, then \(5\) are left, then \(4\), \(3\), \(2\), and \(1\). Multiply, $$6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720,$$ so there are \(\boxed{720}\) orders. Placing a book uses it up, so each slot has one fewer choice than the last. That countdown product is \(6!\).
Problem
\(9\) club members compete for president, vice president, and treasurer. Nobody holds two jobs. How many ways can the three jobs be filled?
Show a hint
- Think of three empty boxes in a row, one for president, one for vice president, and one for treasurer. You are filling them one at a time from the \(9\) members, and someone who takes a job cannot be picked again.
- Count the choices for each box in turn. Any of the \(9\) members could be president. Once that person is chosen, only \(8\) people are left to be vice president, and after that only \(7\) are left to be treasurer. Then multiply the box counts together.
- You only have three boxes, so you stop after three steps and multiply \(9 \times 8 \times 7\).
Show the full solution
Any of the \(9\) members could be president. That leaves \(8\) people for vice president and \(7\) for treasurer, so multiply \(9 \times 8 \times 7\). Since \(9 \times 8 = 72\) and \(72 \times 7 = 504\), the jobs can be filled in \(\boxed{504}\) ways. Three jobs means three factors, so stop the countdown after three instead of running it down to \(1\).
Problem
There are \(900\) three-digit numbers. How many of them do NOT have all three digits the same?
Show a hint
- Instead of chasing the huge pile of numbers you want, try counting the tiny pile you do not want first. How many \(3\)-digit numbers have all three digits the same?
- This is the total-minus-bad move from \(10.1\). Count every \(3\)-digit number that is all the same digit, then take that count away from the \(900\). The ones left over are the numbers that do NOT have all three digits the same.
- The all-same numbers are \(111, 222, 333, 444, 555, 666, 777, 888, 999\), which is \(9\) of them. Now do \(900 - 9\).
Show the full solution
Count the ones you do not want. The \(3\)-digit numbers with all three digits the same are \(111, 222, 333, 444, 555, 666, 777, 888, 999\), which is \(9\) of them. Take those from the \(900\) and you get \(900 - 9 = \boxed{891}\). Every three-digit number is either all-same or not, so subtracting the small group leaves exactly the big one. That is the total-minus-bad move from \(10.1\).
Problem
How many three-digit numbers are there? (The leading digit cannot be \(0\).)
Show a hint
- Draw three empty boxes, one for the hundreds digit, one for the tens digit, and one for the units digit. Then ask how many digits are allowed in each box, and notice that the boxes will not all have the same number of choices.
- Fill the tightest box first. The hundreds digit may be anything except \(0\), so count those choices carefully. The other two boxes have no rule at all, so each can be any digit from \(0\) to \(9\). Once you know the choices in each box, the multiplication principle says you multiply them together.
- The hundreds box has \(9\) choices, the tens box has \(10\), and the units box has \(10\), so you are working out \(9 \times 10 \times 10\).
Show the full solution
Three boxes, hundreds, tens, units. The hundreds box cannot be \(0\), so it has the \(9\) choices \(1\) through \(9\), and the tens and units boxes each take any of \(10\) digits. Multiply, $$9 \times 10 \times 10 = 900,$$ so there are \(\boxed{900}\) three-digit numbers. Total minus bad gets there too. All \(10 \times 10 \times 10 = 1000\) digit strings minus the \(100\) that start with \(0\) is \(900\).
Problem
A 4-digit palindrome reads the same forwards and backwards, and the leading digit is not \(0\). How many 4-digit palindromes are there?
Show a hint
- Draw four slots for the four digits. A palindrome forces some of them to copy others, so you do not really get to choose all four. Which slots are you actually free to pick?
- Only the first digit and the second digit are free choices. Once you pick them, the third digit must copy the second and the fourth must copy the first. So count the ways to fill just those two free slots and multiply.
- The first digit has \(9\) choices (it cannot be \(0\)), and the second digit has \(10\) choices (any of \(0\) through \(9\)). The last two are forced. Multiply \(9 \times 10\) to finish.
Show the full solution
Only the first two digits are free, since the third has to copy the second and the fourth has to copy the first. The first digit has \(9\) choices because it cannot be \(0\), and the second has \(10\). Multiply, \(9 \times 10 = 90\), so there are \(\boxed{90}\) four-digit palindromes. A forced slot has \(1\) way, so it contributes a factor of \(1\) and never changes the product.
Problem
Six people line up. The tallest must stand at the far left and the shortest at the far right. How many orders for the whole row are possible?
Show a hint
- Two of the six spots are already decided for you. The far left end has to be the tallest person and the far right end has to be the shortest person, so those two spots have no choices left. The only real question is what happens in the four spots in the middle.
- Pin down the forced ends first and then forget about them. That leaves four people who still need places and exactly four open middle spots, and every one of those people is different. So you are just arranging four distinct people in four spots in a row.
- Fill the four middle spots one at a time. The first open spot has \(4\) people who could go there, then \(3\) left for the next, then \(2\), then \(1\). Multiply those together, which is \(4 \times 3 \times 2 \times 1 = 4!\), and that is your count.
Show the full solution
The two ends are forced, the tallest on the left and the shortest on the right, so each end has \(1\) way. That leaves \(4\) people for the \(4\) middle spots, and those fill in \(4 \times 3 \times 2 \times 1 = 24\) ways. The whole row has \(\boxed{24}\) orders. Pinning down the forced spots first turns a six-person lineup into a plain arrangement of four.
Problem
A \(4\)-digit PIN has an odd first digit, an even second digit (including \(0\)), and all four digits distinct. How many PINs follow all three rules?
Show a hint
- Draw four empty boxes, one for each digit, and fill them left to right. Settle the digit with the strictest rule first, then move along, and remember that once a digit is used it cannot show up again.
- The first box must be odd, so count how many odd digits there are from \(0\) to \(9\). The second box must be even, so count the even digits. After that the third and fourth boxes can be any leftover digit, so subtract the ones already taken.
- You should have \(5\) odd choices for the first box and \(5\) even choices for the second. The third box is any digit except the two already used, which leaves \(8\), and the fourth box leaves \(7\). Multiply all four counts together.
Show the full solution
The first digit must be odd, so it is one of \(1, 3, 5, 7, 9\), giving \(5\) choices. The second must be even, one of \(0, 2, 4, 6, 8\), another \(5\). The third digit has to dodge the \(2\) digits already used, leaving \(8\), and the fourth dodges \(3\), leaving \(7\). Multiply, $$5 \times 5 \times 8 \times 7 = 1400,$$ so \(\boxed{1400}\) PINs work. Odd and even digits never overlap, so the second box keeps all \(5\) of its choices.
Problem
A locker tag is one letter then three distinct digits, and the first digit cannot be \(0\). How many different tags are possible?
Show a hint
- Picture the tag as four boxes in a row, one for the letter and three for the digits. The slot-and-box habit lets you count one box at a time and then multiply.
- Fill the boxes left to right and count the choices for each, but watch the two catches. The first digit cannot be \(0\), and every later digit has to dodge the digits you already used.
- The letter has \(26\) choices. The first digit has \(9\) choices since \(0\) is banned. The second digit has \(9\) choices, any of the \(10\) digits except the one already placed. The third digit has \(8\) choices. Now find \(26 \times 9 \times 9 \times 8\).
Show the full solution
The letter box has \(26\) choices. The first digit cannot be \(0\), so it has \(9\). The second digit can be any digit except the one already used, so \(9\) again, and the third must dodge both, leaving \(8\). Multiply, \(26 \times 9 = 234\), then \(234 \times 9 = 2106\), then \(2106 \times 8 = 16848\). There are \(\boxed{16848}\) tags. Two different restrictions can land on the same box count for different reasons.
Practice these ideas
Practice
You have \(3\) hats and \(6\) scarves. How many different hat-and-scarf pairs can you make?
Show the solution
One hat pairs with each of the \(6\) scarves, so that hat alone gives \(6\) pairs. All \(3\) hats give three groups of \(6\) with no overlap, so \(3 \times 6 = \boxed{18}\) pairs.
Practice
An ice cream stand has \(5\) cone types and \(5\) flavors (\(2\) fruit and \(3\) chocolate). How many different cones can you order?
Show the solution
There are \(2 + 3 = 5\) flavors in all. Each of the \(5\) cone types pairs with each flavor, so \(5 \times 5 = \boxed{25}\) cones. The flavors arrive in two groups, so add those together first, then multiply across the two stages.
Practice
A breakfast combo is one drink, one main, and one fruit: \(2\) drinks, \(5\) mains, \(3\) fruits. How many different combos are possible?
Show the solution
Multiply the options at each step, \(2\) drinks, \(5\) mains, \(3\) fruits. The first two steps give \(2 \times 5 = 10\) drink-and-main pairs, and each of those takes any of \(3\) fruits, so \(10 \times 3 = \boxed{30}\) combos.
Practice
A signal has \(3\) lights, each glowing in one of \(4\) colors. Repeats allowed. How many different signals are possible?
Show the solution
Repeats are allowed, so each of the \(3\) lights has all \(4\) colors open to it. Multiply, \(4 \times 4 \times 4 = 4^3 = \boxed{64}\) signals. The first light's color does not use anything up, which is why the count stays at \(4\) every time.
Practice
How many \(4\)-digit numbers are there? (The leading digit cannot be \(0\).)
Show the solution
The leading slot cannot be \(0\), so it has the \(9\) choices \(1\) through \(9\). The other three slots each take any digit from \(0\) to \(9\), so \(10\) choices each. Multiply, $$9 \times 10 \times 10 \times 10 = 9000,$$ so there are \(\boxed{9000}\) four-digit numbers.
Practice
How many different orders can \(5\) different trophies be arranged on a shelf?
Show the solution
Fill the five spots left to right. The first has \(5\) trophies to choose from, then \(4\) are left, then \(3\), then \(2\), then \(1\). Multiply, \(5 \times 4 \times 3 \times 2 \times 1 = 120\), so there are \(\boxed{120}\) orders. Each trophy placed is used up, and that countdown to \(1\) is exactly what \(5!\) means.
Practice
A race has \(8\) runners. The fastest gets gold and the next gets silver. How many ways can the two medals be awarded?
Show the solution
Think of two spots to fill in order, the gold first and then the silver. For the gold, any of the \(8\) runners could finish first, so there are \(8\) choices. Once the gold is decided, that runner is taken, which leaves \(7\) runners who could come in second for the silver. The multiplication principle says you multiply the number of choices for each spot, and you get $$8 \times 7 = 56.$$ So the two medals can be awarded in \(\boxed{56}\) different ways.
Practice
A tag has \(2\) characters from \(9\) symbols. Repeats are allowed, but the two characters cannot match. How many tags are possible?
Show the solution
The left spot takes any of the \(9\) symbols. The right spot can be anything except the symbol you just placed, so \(8\) are still allowed. Multiply, \(9 \times 8 = \boxed{72}\) tags. Repeats are allowed in general, and the matching rule bans exactly one symbol on the right, so the count drops by just \(1\).
Practice
Four paintings hang in a row. One special painting is fixed in the far left spot. How many different orders for the whole row are possible?
Show the solution
The star painting is locked into the far left spot, so only the other three move. Fill the three open spots one at a time with \(3\), then \(2\), then \(1\) choices, which gives \(3 \times 2 \times 1 = \boxed{6}\) orders. A locked spot has \(1\) way, so it contributes a factor of \(1\) and drops out of the product.
Practice
A 4-digit palindrome reads the same forwards and backwards, and no digit may be \(0\). How many such palindromes are there?
Show the solution
A 4-digit palindrome has the pattern \(ABBA\), so only the first two digits are free. No digit may be \(0\), so each free digit is one of \(1\) through \(9\), which is \(9\) choices. Multiply, \(9 \times 9 = \boxed{81}\) palindromes. The last two digits are copies, so each contributes a factor of \(1\).
Practice
There are \(900\) three-digit numbers, and exactly \(90\) of them are palindromes. How many three-digit numbers are NOT palindromes?
Show the solution
It is easier to count the ones we do not want and take them away. There are \(900\) three-digit numbers in total, and \(90\) of them are palindromes. So the numbers that are not palindromes are everything else, which gives \(900 - 90 = 810\). So the answer is \(\boxed{810}\).
Practice
There are \(3\) roads from A to B and \(4\) roads from B to C. How many different routes are there from A to C?
Show the solution
Pick a road from A to B in \(3\) ways, then a road from B to C in \(4\) ways. Every first road opens all \(4\) ways to finish, so multiply, \(3 \times 4 = \boxed{12}\) routes.
Practice
A pizza has one each of: \(2\) crusts, \(2\) sauces, \(3\) cheeses, \(4\) toppings. How many different pizzas can you build?
Show the solution
Multiply the options at each step, \(2\) crusts, \(2\) sauces, \(3\) cheeses, \(4\) toppings. Left to right, \(2 \times 2 = 4\), then \(4 \times 3 = 12\), then \(12 \times 4 = 48\), so there are \(\boxed{48}\) pizzas.
Practice
A locker code is \(2\) distinct letters then \(2\) distinct digits. How many different codes are possible?
Show the solution
The first letter has \(26\) choices and the second must differ, leaving \(25\), so the letters give \(26 \times 25 = 650\). The first digit has \(10\) choices and the second \(9\), so the digits give \(90\). Then \(650 \times 90 = \boxed{58500}\) codes. Making a pair distinct only trims the second slot by one, and everything still multiplies.