Counting

Counting may have been one of humankindÂ’â€™s first thought processes; nevertheless, counting can be deceptively hard. In part, because we often forget some of the principles of counting, but also because counting can be inherently difficult.

When counting elements that are in overlapping sets, the total number will equal the number in one group plus the number in the other group minus the number common to both groups. Venn diagrams are very helpful with these problems.

Example 1:

If in a certain school 20 students are taking math and 10 are taking history and 7 are taking both, how many students are taking either math or history?

(A) 20

(B) 22

(C) 23

(D) 25

(E) 29

Solution:

By the principle stated above, we add 10 and 20 and then subtract 7 from the result. Thus, there are (10 + 20) Â–â€“ 7 = 23 students.

The number of integers between two integers inclusive is one more than their difference.

Example 2:

How many integers are there between 49 and 101, inclusive?

(A) 50

(B) 51

(C) 52

(D) 53

(E) 54

By the principle stated above, the number of integers between 49 and 101 inclusive is (101 Â–â€“ 49) + 1 = 53.

To see this more clearly, choose smaller numbers, say, 9 and 11. The difference between 9 and 11 is 2. But there are three numbers between them inclusiveâ€”9, 10, and 11â€”one more than their difference.

Fundamental Principle of Counting: If an event occurs m times, and each of the m events is followed by a second event which occurs k times, then the second event follows the first event mÂ·k times.

The following diagram illustrates the fundamental principle of counting for an event that occurs 3 times with each occurrence being followed by a second event that occurs 2 times for a total of 3Â·2 = 6 events:

Example 3:

A drum contains 3 to 5 jars each of which contains 30 to 40 marbles. If 10 percent of the marbles are flawed, what is the greatest possible number of flawed marbles in the drum?

(A) 51

(B) 40

(C) 30

(D) 20

(E) 12

There is at most 5 jars each of which contains at most 40 marbles; so by the fundamental counting principle, there is at most 5Â·40 = 200 marbles in the drum. Since 10 percent of the marbles are flawed, there is at most 20 = 10% Â·200 flawed marbles.

Example 1:

In a legislative body of 200 people, the number of Democrats is 50 less than 4 times the number   of Republicans. If one fifth of the legislators are neither Republican nor Democrat, how many of     the legislators are Republicans?

(A) 42

(B) 50

(C) 71

(D) 95

(E) 124

Solution

Let D be the number of Democrats and let R be the number of Republicans. "One fifth of the legislators are neither Republican nor Democrat," so there are $\frac{1}{5}$ Â·200 = 40 legislators who are neither Republican nor Democrat. Hence, there are 200 Â–â€“ 40 = 160 Democrats and Republicans, or D + R = 160. Translating the clause "the number of Democrats is 50 less than 4 times the number of Republicans" into an equation yields D = 4R Â–â€“ 50. Plugging this into the equation D + R = 160 yields

4R Â–â€“ 50 + R = 160

5R Â–â€“ 50 = 160

5R = 210

R = 42

Example 2:

Speed bumps are being placed at 20 foot intervals along a road 1015 feet long. If the first speed   bump is placed at one end of the road, how many speed bumps are needed?

(A) 49

(B) 50

(C) 51

(D) 52

(E) 53

Solution

Since the road is 1015 feet long and the speed bumps are 20 feet apart, there are 1015 20  = 50.75 , or 50 full sections in the road. If we ignore the first speed bump and associate the speed bump at the end of each section with that section, then there are 50 speed bumps (one for each of the fifty full sections). Counting the first speed bump gives a total of 51 speed bumps. The answer is (C).