Topic 1: The Fundamental Counting Principle

Counting MethodsTopic 4: Permutations with RestrictionsI can generalize strategies for determining the number of permutations when restrictions are involved.Example 1In how many different ways can the letters in the word COMPARE be arranged:

without restrictionsbeginning with two consonants?beginning with exactly two consonants, that is, the first two letters are consonants and the third letter is not a consonant?alternating consonants and vowels?M is not the first letter

Arranging letters with restrictionsExample 1: SolutionCOMPAREwithout restrictions?

beginning with two consonants?

__ __ __ __ __ __ __C C

Once youve used a consonant, there are only 3 left to choose from for the second space.

Since you have now used 2 letters, there are 5 letters left to put in the 5 remaining spaces.

There are 4 possibilities for the first space, since there are 4 consonants in the letters in the word COMPARE.Example 1: SolutionCOMPARE

beginning with exactly two consonants, that is, the first two letters are consonants and the third letter is not a consonant?

__ __ __ __ __ __ __C C C

Once youve used a consonant, there are only 3 left to choose from for the second space.

The third letter cannot be a consonant. Therefore, there are 3 possibilities for the 3rd space.

There are 4 possibilities for the first space, since there are 4 consonants in the word COMPARE.

There are 4 letters left to place in the 4 remaining spaces.Example 1: SolutionCOMPARE

alternating consonants and vowels?

__ __ __ __ __ __ __C V C V C V C

There are 4 consonants to be placed in 4 spaces.

There are 3 vowels to be placed in 3 vowel spaces.

There are 4 consonants and 3 vowels. In order for them to alternate, a consonant has to come first.Example 1: SolutionCOMPARE

M is not the first letter

__ __ __ __ __ __ __M

There are 6 letters that are not M. Any of these 6 letters could be the first letter. Example 2A family consisting of 5 people, 2 adults and 3 children, are going to a movie. In how many ways can they sit in 5 adjacent seats under the following conditions?

a) the parents must sit togetherb) the children must sit togetherc) the children must sit together and the adults must sit togetherd) the father must not sit in the middle

Arranging objects with restrictionsExample 2: Solution2 adults and 3 children

a) the parents must sit togetherWe treat the parents as a group with 2! arrangements.Now there are 4 things to arrange: 3 children and 1 group of adults.

Within one of the things we are arranging (the adult group), there are 2! arrangements.There are 4 things to arrange (the 3 children and the 1 adult group).Example 2: Solution2 adults and 3 children

b) the children must sit together

We treat the children as a group with 3! arrangements.Now there are 3 things to arrange: 2 adults and the 1 group of children.

Within one of the things we are arranging (the children group), there are 3! arrangements.There are 3 things to arrange (the 2 adults and the 1 group of children).Example 2: Solution2 adults and 3 children

c) the children must sit together and the adults must sit together

We treat the parents as a group with 2! arrangements and the children as a group with 3! arrangements.Now there are 2 things to arrange: 1 group of children and 1 group of adults.

Within one of the things we are arranging (the adult group), there are 2! arrangements and within the other group (the children group) there are 3! arrangements.There are 2 things to arrange (1 group of children and 1 group of adults).Example 2: Solution2 adults and 3 children

d) the father must not sit in the middle

___ ___ ___ ___ ___F

There are 4 other family members who can sit in the middle.Example 3At a used car lot, 9 different car models (3 of which are red) are to be parked close to the street for easy viewing. How many ways can the 9 cars be parked if:

the 9 cars must be parked so that there is a red car at each end and the third red car is exactly in the middle?

all three red cars must be parked side by side?

the three red cars must not all be parked side by side?

Arranging objects with restrictionsR R RExample 3a) the 9 cars must be parked so that there is a red car at each end and the third red car is exactly in the middle?b) the three red cars must be parked side by side?

__ __ __ __ __ __ __ __ __

Once youve placed the 1st red car, there are 2 possibilities for the next red car and 1 for the next.

The 6 other cars are now placed in the remaining six spaces.

There are 3 possibilities for the first space, since there are 3 red cars.

We treat the 3 red cars as a group with 3! arrangements.Now there are 7 things to arrange: 6 non-red cars and 1 red car group.

Within the things group of red cars, there are 3! arrangements.There are 7 things to arrange.Example 3c) the three red cars must not all be parked side by side?In order to find the number of ways in which the three cars must not ALL be parked side by side,Determine the total number of ways to arrange the carsSubtract the number of ways in which the cars would be all three parked together

Example 4Using only the digits 0, 1, 2, 3, 4, and 5, without repetitions, how many natural numbers are:

6-digit natural numbers? 6-digit natural odds numbers?6-digit natural numbers with alternating even and odd digits?

Arranging digits with restrictionsExample 4: Solutiona) 6-digit natural numbers? 0, 1, 2, 3, 4, 5b) 6-digit natural odd numbers?odd1,3,5__ __ __ __ __ __ There are 3 odd numbers that can be the last digit. 2 cases will have to be used. No repetition and a 6-digit natural number cannot start with a 0 or it would be a 5-digit natural number.__ __ __ __ __ __ 0

1) the first digit is an odd number and the last digit is odd2) the first digit is an even but not 0.__ __ __ __ __ __Odd but already used 1 at the startOdd1,3,5Even not 0 2,4or

+144 + 144 = 288Example 4: Solution0, 1, 2, 3, 4, 5c) 6-digit natural numbers with alternating even and odd digits?O E O E O E __ __ __ __ __ __ There are 3 odd numbers and 3 even numbers, remember that 0 cannot be the first digit when staring with even numbers. 2 cases will have to be used. 1) Starting with odd2) the first digit is an even but not 0.__ __ __ __ __ __ Even not 0 2,4or

+36 + 24 = 60E O E O E O Need to KnowBoth the fundamental counting principle and factorial notation can be used to solve permutations involving restrictions.

Restrictions should be addressed first.

Need to KnowTo find the number of arrangements of n objects, with a group of k adjacent objects, treat the group of adjacent objects as one object and multiply the number of arrangements of the objects by the number of arrangements within the group of adjacent objects:

(number of arrangements of the objects) x (number of arrangements within group)

Need to KnowTo find the number of arrangements that do not satisfy a restriction calculate:

(total number of arrangements) (number of arrangements that do satisfy the restriction)

Youre ready! Try the homework from this section.