**Combinations/Probability**

This page gives examples of combination, permutation, and probability problems.

**Combination** problems and **Permutation** problems are generally solved through some form of calculation, which is usually be done by multiplication. We split the problem into separate parts, calculate the number of instances in each part, and then combine the results of our calculation.

A permutation is an arrangement of a set of objects in a particular order. Permutation problems explore the possible arrangements and number of arrangements of objects selected from the set. They may also involve arrangements where some of the objects are repeated, or where there are multiple sets, each individually with their own permutations, but then combined in a single arrangement. The basic formula for the number of permutations of p things selected from a set of size n is:

\(\begin{align} \frac {n!}{(n-p)!}\end{align}\)

Combinations are selections from sets of objects where the order in which the selection is arranged is unimportant. The basic formula for the number of combinations of p things selected from a set of size n is:

\(\begin{align} \frac {n!}{(n-p)!p!}\end{align}\)

\(\begin{align} n \choose p\end{align}\)

which is said, “n choose p.”

**Probability** problems are solved at the basic level by using the simple formula:

\(\begin{align}probability = \frac {number\hspace{1mm} of\hspace{1mm} favorable\hspace{1mm} outcomes}{number\hspace{1mm} of\hspace{1mm} possible\hspace{1mm} outcomes}\end{align}\)

Since both the numerator and denominator of this fraction are often determined through some form of “counting,” then *probability* problems often are solved by solving *permutation* and *combination* problems first.

On this page we have 1 Permutations, 3 Combinations, and 2 Probability examples.

**To practice more of these types of problems, click** here.

**Permutations Example**

The school orchestra’s string section has 4 chairs designated for violins, 2 for violas, 2 for cellos, and 2 for basses, with the corresponding number of students playing each of these instruments. The conductor wants the students to sit in a different, randomly chosen order each day in practice. How many days can they do this before they repeat a seating order?

A. 32

B. 48

C. 96

D. 128

E. 192

**Explanation:**

This is a straightforward permutations problem involving 4 different permutations – 1 for each instrument section. Therefore, the solution is the number of permutations of seating order for the violins, times that for violas, times that for cellos, times that for basses. Or

4! 2! 2! 2! = (4 × 3 × 2 × 1) × (2 × 1) × (2 × 1) × (2 × 1) = 24 × 2 × 2 × 2 = 192

**So E is the correct answer.**

**Combinations Example 1)**

A state’s license plates consists of 2 letters, followed by 4 numeric digits, with these restrictions: (1) Two letters, O and I, are never used, and (2) 0 is never the first of the 4 digits. How many possible license plates are there for the state?

A. 5,189,184

B. 5,616,000

C. 5,630,625

D. 5,184,000

E. 5,183,424

**Explanation:**

This is a combination problem whose answer can be calculated by multiplying the correct numbers.

1. We need to know how many different combinations of two letters are possible if the letters O and I are never used. This means there are 24 letters available, and so the number of combinations of letters is 24 x 24 = 576.

2. Next, we need to know how many 4 digit numbers there are if we disallow 0 from being the first digit. What this means is that we want to know how many numbers there are between 1000 and 9999, inclusive. This is determined by subtracting 999 from 9999. The result is 9000.

3. Finally, we multiply the results of 1 and 2: 576 × 9000 = 5,184,000.

**So D is the correct answer.**

**Combinations Example 2)**

If a combination lock has 40 numbers on it, and each combination for unlocking the lock uses 3 of the

numbers with repetitions allowed, then how many different combinations are possible?

A. 24,000

B. 44,000

C. 82,000

D. 64,000

E. 36,000

Explanation: The order of the numbers in a combination is important, and each number can be repeated

in a combination 1 or 2 times. So, we have 40 choices for the first number, 40 for the second, and 40 for

the third. The total number of possible combinations then is

40 × 40 × 40 = 64,000

**So D is the correct answer.**

**Combinations Example 3)**

If a combination lock has x numbers on it, where x is a multiple of 10, and each combination for unlocking the lock uses 3 of the numbers with repetitions allowed, then what is the smallest possible value for x if it’s desired to have at least 10,000 different combinations?

A. 33

B. 30

C. 20

D. 27

E. 10

Explanation: The number of possible combinations is x\(\begin{align}^3\end{align}\). Since x is a multiple of 10, answers A and D are immediately eliminated. Also, we want the minimum value of x that will yield 10,000 combinations, so we’ll try ascending values for x on the other 3 possible answers:

x = 10: \(\begin{align}x^3\end{align}\) = 1000

x = 20: \(\begin{align}x^3\end{align}\) = 8000

x = 30: \(\begin{align}x^3\end{align}\) = 27,000

Obviously, 30 is the desired number, **so B is the correct answer.**

**To practice more of these types of problems, click** here.

**Probability Example 1)**

When two dice are rolled, the probability that 7 will come up is 1/6, that 4 will come up is 1/12, and that 9 will come up is 1/9. What is the probability that in three rolls of the dice, the first will be 4, the second will not be 9, and the third will not be 7?

A. 5/81

B. 1/8

C. 4/27

D. 5/13

E. 3/16

Explanation: Because each of these rolls is independent of the others, to get the probability that **all three** events will happen, we need to **multiply** the following 3 probabilities:

4 Probability = 1/12

not 9 Probability = 1 – 1/9 = 8/9

not 7 Probability = 1 – 1/6 = 5/6

1/12 × 8/9 × 5/6 = 5/81

**So A is the correct answer.**

**Probability Example 2)**

In a certain neighborhood the probability that person’s birthday is not in January is .9, and not in July is .85. What’s the probability that a person selected at random from the neighborhood is born in either January or July?

A. 1/9

B. 1/8

C. 1/4

D. 1/6

E. 1/3

Explanation: The probability of an event’s happening is 1 minus the probability of its *not* happening. Therefore, the probability of a person in the neighborhood being born in January is .1, and in July is .15. Since no one can be born in both months, then the probability of someone being born in either month is additive. So the probability of being born in either January or July is

.1 + .15 = .25 = 1/4

**So C is the correct answer.**

**To practice more of these types of problems, click** here.