College Algebra Tutorial 55

College Algebra
Answer/Discussion to Practice Problems
Tutorial 55: Fundamental Counting Principle

WTAMU > Virtual Math Lab > College Algebra > Tutorial 55: Fundamental Counting Principle

Answer/Discussion to 1a

One quarter, one dime and one six-sided die are tossed. How many results are possible?

Let's use the basic counting principle:

There are 3 stages or events: one quarter, one dime, and one six-sided die.

Each coin has 2 possible outcomes, either a tail or a head.

The die has 6 possible outcomes.

Putting that all together we get:
quarter dime die # of possible outcomes 2 x 2 x 6 = 24

So there are 24 different possible outcomes.

(return to problem 1a)

Answer/Discussion to 1b

Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?

Let's use the basic counting principle:

There are 4 stages or events: a science class, a math class, a history class and an english class.

There are 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to pick from.

Putting that all together we get:
Science Math History English # of schedules 4 x 3 x 2 x 3 = 72

So there are 72 different schedules possible.

(return to problem 1b)

Answer/Discussion to 1c

Six students in a speech class all have to give there speech on the same day. One of the students insists on being first. If this student's request is granted, how many different ways are there to schedule the speeches?

Let's use the basic counting principle:

There are 6 stages or events: speaker 1, speaker 2, speaker 3, speaker 4, speaker 5, and speaker 6.

There is only one possibility for speaker 1.

That leaves 5 possibilities for speaker 2, which leaves 4 for speaker 3, which leaves 3 for speaker 4, which leaves 2 for speaker 5 which leaves 1 for speaker 6.

Putting that all together we get:
speaker 1 speaker 2 speaker 3 speaker 4 speaker 5 speaker 6 # of ways

1 x 5 x 4 x 3 x 2 x 1 = 120

So there are 120 different ways they can be scheduled to speak.

(return to problem 1c)

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