Counting and Probability
Counting Problems
Counting problems ask you how many ways an event can occur or how many possible
outcomes there are to a situation. Specifically, counting problems often deal with
permutations (order matters) or combinations (order does not matter). Combinations come up
only on rare occasion on the ACT, and somewhat more frequently on the SAT. On any counting
problem, you should write out one “slot” for each event that is occurring, populate these slots
with the number of ways each event can occur, and multiply these numbers together, as is
demonstrated in the examples below.
1) A cafeteria offers students a choice of one of three entrees, one of two salads, and one of
four desserts. How many different meals consisting of one entrée, one salad, and one desert
can a student create?
Three events are occurring (an entrée, a salad, and a dessert are each chosen). Therefore, you
should draw out three “slots” with multiplication signs in between them. __ • __ • __
Populate each slot with how many ways that event can occur: 3 • 2 • 4 = 24 different possible
meals.
2) How many ways can five students arrange themselves in a line?
Five events are occurring (one for each spot in the line being filled). Therefore, you should
draw out five “slots” with multiplication signs in between them. __ • __ • __ • __ • __
Populate each slot with how many ways that event can occur. There are five options for the
first spot in line. No matter which student is chosen for the first spot in line, only four options
are left for the second spot. By the same logic, the third spot has three options, the fourth spot
has two options, and the fifth spot has only one option. Therefore, the slots can be populated
as follows: 5 • 4 • 3 • 2 • 1 = 120 ways the students can arrange themselves in line. 5 • 4 • 3 •
2 • 1 is also known as 5! (pronounced “5 factorial”). The factorial function on your TI calculator
can be found under Math Probability
3) How many four digit numbers can be formed using only odd digits?
Four events are occurring because a digit must be chosen for each of the four digits in the four
digit number. Therefore, you have __ • __ • __ • __. Because there are five odd digits, 1, 3, 5,
7, and 9, there are five options for each of four events occurring, so you have 5 • 5 • 5 • 5 = 54 =
625.
4) How many four digit numbers contain the digits 5, 6, 7, and 8?
At first, this problem seems very similar to problem 3. However, problem 3 did not specify that
numbers could not be chosen twice, so four digit numbers like 5,753 or 7,777 were acceptable
outcomes and had to be counted. In this problem, however, it is specified that the number
must contain the digits 5, 6, 7, and 8. In order for a four digit number to contain each of four
unique digits, it can only use each digit once. Therefore, the four events that are occurring, __
• __ • __ • __, can be populated as follows: 4 • 3 • 2 • 1 = 24 four digit numbers that contain
the digits 5, 6, 7, and 8. This question is in fact much more similar to question 2 than question
3.
5) There are 50 people entered in a raffle in which the grand prize winner wins a car and the
second prize winner wins a bike. How many possible outcomes are there for the raffle,
assuming that one person cannot win both prizes?
Two events are occurring, so __ • __. There are 50 possible grand prize winners, leaving only
49 possible winners for the second prize because the same person cannot win both prizes, as
specified in the problem. Therefore, you have 50 • 49 = 2450 possible outcomes.
6) There are 50 people entered in a raffle in which two grand prize winners will each win
identical cars. How many possible outcomes are there for the raffle, assuming that one person
cannot win both prizes?
Although this problem might look very similar to problem 5, it is actually quite different.
Problem 5 was a permutation, since order mattered. It would be a very different outcome if
Tommy won the car and Suzy won the bike than if Suzy won the car and Tommy won the bike.
This problem, on the other hand, is a combination because order does not matter. If Tommy
and Suzy win, the outcome is settled, as they both win identical cars. Begin by calculating 50 •
49. However, as is demonstrated in problem 5, this calculation assumes that order does
matter, which is incorrect in this situation. To select out for overlap, you must divide by the
number of ways the two winners can be arranged among themselves: 2! Therefore, you have
(50 • 49)/2! = 1,225.
7) Three student council members are to be chosen from a class of 100 students. How many
combinations of students are possible for the three council members?
This problem, like question 6, is a combination problem. Order is unimportant because each
seat on student council is identical. By the reasoning described in question 6, you have (100 •
99 • 98) / 3! = 161,700.
Probability
The general formula for probability is (# of desired outcomes) / (# of possible outcomes). The
possible outcomes are usually figured out by using the processes described above for counting
problems. The desired outcomes are sometimes figured out using these same processes and
are sometimes more easily counted manually. The probability of consecutive events occurring
is found by multiplying their individual probabilities.
8) A dice is rolled. What is the probability that a 6 is rolled?
Because a dice has 6 sides, there is 1 possible outcome and 6 desired outcomes, so the
probability is 1/6.
9) Two dice are rolled. What is the probability that two 6’s are rolled?
When two dice are rolled, there are 6 ways the first can land and 6 ways the second can land,
so there are a total of 6 • 6 = 36 possible outcomes. There is only one desired outcome (both
rolls are 6’s), so the probability is 1/36. Another way of thinking about it is that there is a 1/6
probability of the first roll being a 6 and a 1/6 probability of the second roll being a 6, so the
probability of both being 6’s is 1/6 • 1/6 = 1/36.
10) Two dice are rolled. What is the probability that a one lands a 5 and the other lands a 6?
As established in question 9, there are 36 possible outcomes. In this problem, there are two
possible outcomes: dice A is a 5 and dice B is a 6, or dice A is a 6 and dice B is a 5. Therefore,
the probability of this event is 2/36 = 1/18. Another way of thinking about it is that the
probability of a desired outcome on the first roll is 2/6 since a 6 or a 5 would be desired. Either
way, there is one desired outcome for the second roll, since a 5 must be paired with a 6 and a 6
must be paired with a 5, so the probability of a desired outcome on the second roll is 1/6.
Therefore, the overall probability of getting one 5 and one 6 is 2/6 • 1/6 = 2/36 = 1/18.
11) You roll two dice. What is the probability that the sum of the two numbers rolled is greater
than 8?
As established in question 9, there are 36 possible outcomes. The easiest way to find the
number of desired outcomes (the outcomes with a sum greater than 8) is to count them by
hand: (1) 6, 6; (2) 6, 5; (3) 6, 4; (4) 6, 3; (5) 5, 6; (6) 5, 5; (7) 5, 4; (8) 4, 6; (9) 4, 5; (10) 3, 6. Since
there are 10 desired outcomes, the probability is 10/36 = 5/18.
12) You flip a coin four times. What is the probability that you get exactly 3 tails?
Here you have four events occurring, each of which can occur two possible ways. Therefore,
the possible outcomes are as follows: 2 • 2 • 2 • 2 = 24 = 16. The easiest way to find the
number of desired outcomes (the number of ways you could get exactly 3 tails) is to count
them manually. There are 4 desired outcomes: ttth, ttht, thtt, httt. Therefore, the probability
is 4/16 = 1/4.
13) There are 8 red marbles, 10 white marbles, and 4 blue marbles in a jar. What is the
probability that you draw two red marbles in a row without replacing the marble after your first
draw?
The probability of the first draw being a red is 8/22. To calculate the probability of the second
draw being a red given that the first draw was a red, you must subtract a red marble from the
number of red marbles and the total number of marbles in the jar, giving you 7/21. Therefore,
the probability of drawing two red marbles in a row without replacement is 8/22 • 7/21 =
56/462 = 4/33.