Data



Counting

The Multiplication Principal:

Suppose that there are k choices to be made, with n1 ways to make choice one.
For each of the above choices there are n2 ways to make choice two.
For each of the above choices there are n3 ways to make choice three, and so on and so forth.
Then there are n1*n2* ...*nm ways to perform all k of the operations in sequence.

Apply the Multiplication Principal when the desired object requires an item from each one of the available categories.   If you can reduce the problem down to the word AND, the Multiplication Principal will apply.   

 

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Example 1:

Ken has 6 different pairs of slacks, 8 different shirts, 5 different pairs of shoes, and 3 different ties.    How many outfits consisting of one pair of slacks, one shirt, one pair of shoes, and one tie can he create?

 

Solutions:
Since the object he wants to create, which is an outfit, requires an item from the slacks category, AND the shirt category, AND the shoes category, AND the tie category, the multiplication principal is used.

(6)(8)(5)(3) = 720 Outfits

Therefore, Ken has 720 different outfits available to him.

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Example 2:

At a veggie pizza stand, a pizza can be ordered with any combinations of 8 toppings: tomatoes, peppers, olives, pineapple, artichoke hearts, onion, jalapino, and zucchini.    How many different ways are there to order a pizza and toppings?

 

Solution:
Since there are 8 toppings, there are eight separate choices to be made.     There are eight choices, so there are eight slots.   The first slot is for the choice of tomatoes, the second slot is for the choice of peppers and so on and so forth.

Since we need a choice in the first slot AND the second slot AND the third slot, and so on and so fourth, the operation of multiplication is used.

Set up the following template.

__*__*__*__*__*__*__*__

There are two ways to order each topping:   with or without.   Thus each slot has a choice of 2.

2 * 2 * 2 * 2 * 2 * 2 * 2 * 2

Therefore there 256 different ways to order a Pizza from the veggie stand.

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Example 3:

An automobile license plate consists of three letters (A-Z) followed by three digits (0-9).    How many different license plates are possible?

 

Solution:
To determine if the multiplication rule is applied, ask the following question.

Is it the case where we want a letter in the first slot and a letter in the second slot and a letter in the third slot, etc.

Or is it the case where we want a letter in the first slot or a letter in the second slot or a letter in the third slot, etc.

It is the case of "AND", so the multiplication rule is used.

Set up the following slot problem for three letters and three digits.

__*__*__*__*__*__

The first slot is for a letter.   There are 26 letters in the alphabet, so there are 26 choices for the first slot.   The next two slots are also for letters, so those slots have 26 choices each.

26 * 26 * 26 * __ * __ * __

The fourth slot is for a digit.   There are 10 numbers between zero and nine, so there are 10 choices for the fourth slot.   The last two slots are also for digits, so those slots have 10 choices each.

26 * 26 * 26 * 10 * 10 * 10




17,576,000


Therefore there are 17,576,000 different possible license plates that can be make with the above conditions.

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Example 4:

The same problem as example 3, but with no redundancies.

 

Solution:
No redundancies means that once a letter or a digit has been used, it can not be used again.


There are still six slot.   The first slot must be a letter, and there are 26 ways to place a letter in the first slot.   The second slot must also be a letter, but one of the letters was used on the first slot, and since repetition is not permitted there are 25 ways to place a letter in the second slot.   The third slot must also be a letter, and there are now 24 ways to place a different letter in the third slot.   The same idea applies for the digits.

26 * 25 * 24 * 10 * 9 * 8



11,232,000



Therefore there are 11,232,000 different license plates available with no redundancies.

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The Addition Principal:

Suppose there are k choices to be make.
Let n1, n2, n3, ..., nm be disjoint sets.

Then the number of ways to perform all k of the operations is:
n1 + n2 + n3 + ... + nm


The Addition Principle is used when the object you want to create requires just one item from ONE of the available categories.   If you can reduce the problem down to the word OR, then the Addition Principle applies.   Lets look at the next example.

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Example 5:

Susan wants to buy a birthday present for her boyfriend.   She is considering 5 different CDs, 3 different sweaters, and 6 different shirts.   From these possibilities, how many different presents does she have to chose from?

Solution:
Since the object that Susan wants to create is a birthday present, and that present consists of just ONE item from ONE of the choices the addition principle is used.   You can also think of it as an OR problem.   Susan wants a CD, OR a sweater, OR a shirt.   The word OR means add.

5 + 3 + 6 = 14

Therefore Susan has 14 possible choices for the birthday present.

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Think about it!

How would the above problem need to be worded so that the multiplication is used instead of the addition principal?






The present would have to consist of a CD AND a sweater, AND a shirt.

Susan wants to buy a birthday present for her boyfriend.   The present will consist of a CD, a sweater, and a shirt.   She is considering 5 different CDs, 3 different sweaters, and 6 different shirts.   From these possibilities, how many different presents does she have to chose from?



Since the present now requires an object from each category: CD, sweater, and shirt, the Multiplication Rule is used.

(5)(3)(6) = 90

Therefore there are 90 different possible presents.





Student #1:

A math student has job offers from 4 insurance companies with headquarters in Los Angeles, 2 companies with headquarters in New York, and 3 companies with headquarters in San Francisco.   How many different choices of companies are available to her?


9               24               200               96              


Answer



Example 6:

A committee is to elect a president, secretary, and treasurer.   If the committee consists of 45 members, how many outcomes for the three positions can occur?
(No person can win two positions)

Solution:
To create the three positions, we will need someone in each of the categories of president, secretary, and treasurer.   This will require the Multiplication Principle.   There are three choices to make so there are three slots.
__ * __ * __

There are 45 ways to place a person in the first slot.   Once a person is placed in the first slot, there are only 44 people left in which to place a person in the second slot.   There are now 43 ways to place a different person in the third slot.

45 * 44 * 43

85,140

Therefore there are 85,140 different ways to create the committee.





Student #2:

A three digit number is to be formed using the numbers 2 through 8.   How many different numbers can be formed without repetition?


30               210               7050               3,024              


Answer



Example 7:

 A factory makes special order rings with the clients initials engraved upon the top of the rings.   Rings are made with either 2 initials or 3 initials.   How many rings would the factory need to have in stock in order to meet any clients request?

Solution:
The requirement is as follows:
A ring can be a letter in the first slot AND a letter in the second slot, OR the ring can be a letter in the first slot AND a letter in the second slot AND a letter in the third slot.

The word AND means to multiply and the word OR means to add.   The template is as follows:

__ * __   +   __ * __ * ___


The first two slots are the number of ways a two initial ring can be made, and the last three slots are the number of ways a three initial ring can be made.

There are 26 possible choices for each slot.

26 * 26   +   26 * 26 * 26



Therefore the factory must have 18,252 rings in order to meet any request.





Student #3:

Same problem as example 7, except without repe2ition.
In other words, once you use a letter it may not be used again.

12,000               11,900               16,250               8,790              


Answer



Example 8:

In how many different orders can 3 goats and 2 cows be lined up under the following conditions.

A)   The first in line must be a goat.

Solution:
Since we need an animal in the first slot AND the second slot AND the third slot, etc., the multiplication property is used.   Five animals are lined up so there are five slots.

__ * __ * __ * __ * __

The condition is that the first animal in line must be a goat.   There are 3 goats, so there are three ways to place a goat in the first slot.   Since a goat has just been placed in the first slot we are down to 2 goats and 2 cows.   Therefore there are 4 animals left, and thus 4 choices to place an animal in the second slot.   There are now 3 animals left, and thus there are 3 choices to place an animal in the third slot, etc.

3 * 4 * 3 * 2 * 1 = 72

There are 72 ways to line up the animals in which a goat is first in line.

B)   The one in the middle and the end of the line must be a cow.

Solution:
Again there are five slots.

__ * __ * __ * __ * __

The condition states that the animal in the middle and at the end of the line must be a cow.
There are 2 cows, so there are 2 ways to place a cow in the middle slot.   Then the last cow will have to be placed in the last slot.   There is just 1 way to place the last cow in the last slot.

__ * __ *   2   * __ * 1

There are three animals remaining which are goats, so there are 3 ways to place a goat in the first slot, and then 2 ways to place a goat in the second slot, and then 1 way to place the last goat in the fourth slot.


3 * 2 * 2 * 1 * 1

12

Therefore there are 12 different ways to line up the animals with a cow in the middle and at the end of the line.

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Example 9:

At a small farmers fair there are 12 goats, 15 pigs, and 10 sheep.   If two awards are to be given find the following:

 

The two awards are different.

A)   In how many different ways can the awards be given if the awards can be given to any animal? (The same animal may win both awards)

Solution:
Since the first award is to given out AND the second award is to be given, the multiplication principal is used.   Thus there are two slots.

There are 37 animals and each award may be given to any animal.

37 * 37

1369   ways

 

B)   In how many different ways can the awards be presented if they have to be given to different animals?

Solution:
There are 37 different ways to assign the first award.   The animal that wins the first award can not win the second award, so that leaves 36 different ways to give out the second award.

37 * 36

1332   ways

 

C)   In how many different ways can the awards be given to animals of different breeds?

Remember the awards are different!

There are:   12 goats, 15 pigs and 10 sheep

Since the awards must be given to animals of different breeds, giving the first award to a goat and the second award to a pig would satisfy the criteria.   What else would satisfy the condition of giving the two different awards to different breeds? Think about it before you go on!

 

Let   g = goat, p = pig, and s = sheep.

Awards can be assigned as follows:

g and p    or    g and s    or    p and s

(12)(15) + (12)(10) + (15)(10)

Suppose the first award is for good teeth, and the second award is for a good coat.   Is the following a different situation?

The goat receives the good teeth award and the pig receives the good coat award OR the pig receives the good teeth award and the goat receives the good coat award.

The answer is yes, they are different situations, so both situations must be considered.

Since the awards are different, a goat and a pig, is different than a pig and a goat.

g,p or g,s or p,s    (12)(15)+(12)(10)+(15)(10)

or    p,g or s,g or s,p    (15)(12)+(10)(12)+(10)(15)

Total = 900


Therefore there are 900 ways to give the two different awards to animals of different breeds.

 

The awards are the same award.

D)   In how many different ways can the awards be given to animals of different breeds?

Solution:
Suppose the first award is for good teeth, and the second award is also for good teeth.   Would the following be a different situation?

The goat receives the good teeth award and the pig receives the good teeth award OR the pig receives the good teeth award and the goat receives the good teeth award.

The answer is no, they are not different, so both situations do not need to be considered.

Since the awards are the same:

(15)(12)+(10)(12)+(10)(15) = 450 ways


Therefore there are 450 ways to give the two identical awards to animals of different breeds.





Student #4:
Two free dinners are to be given away at Ron's Cafe.   The two dinners are identical.   There were 5 teacher, 7 construction workers, and 3 postal carriers that registered for the dinners.   In how many different ways can the two dinners be given to people who work in different professions?


71               105               15               168              

Answer

Example 10:

A high school basketball player has been offered scholarships at 4 California universities, 2 Texas universities, 7 New York universities, and 3 Arizona universities.   How many choices of scholarships are possible?

Solution:
Since the player can only choose one of the scholarships, the Addition Principal is used.   In other words, the player will choose one from the state of California, OR one from the state of Texas, OR etc.

4 + 2 + 7 + 3 = 16   choices

 

Therefore there are 16 choices available to the student.





Homework Problems



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Student #1:    The answer is A:

Solution:
The student must choose just ONE job, so the Addition Principal applies.   The student will take a job in Los Angeles, OR New York, OR San Francisco.   The word OR means to add.  

4 + 2 + 3 = 9

Therefore there are 9 different choices available to the student.

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Student #2:    The answer is B:


There are three slots.     __ __ __

The first slot has a choice of 7 numbers.    Once a number has been used in the first slot, there are only 6 choices left for slot two, and then 5 choices left for slot three.


Therefore there are 210 different ways.
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Student #3:    The answer is C:

There are 26 letters in the alphabet.    There is 26 possible ways to place a letter in the first slot, and 25 choices for the second slot.
After the plus sign it is a new problem.   Start with 26 possible choices for the first slot, 25 choices for the second slot, and 24 choices for the third slot.



Therefore there are 16,250 different possible rings.
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Student #4:    The answer is A:

To give the two dinners to people of different professions the possible choices are:
Teacher and Construction worker, OR Teacher and Postal Carrier, Or Construction worker and Postal Carrier.

The word and means multiply, and the word or means add.

__ __ + __ __ + __ __


(5)(7) + (5)(3) + (7)(3) = 71    possible outcomes


Therefore there are 71 different ways to give the two dinners to people of different professions.

Since the dinners are the same, this problem does not need to be multiplied by two.    If the dinners were different, then the solution would be   2(71) = 142.
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