Answer:
1.) 5 choices × 5 choices × 5 choices = 53 possible 3 digit codes including repeats
2.)Choose two letters from four = P(4,2) = 4!/2! = 4 choices × 3 choices = 12. Order matters
i.e. DA is a different word than AD.
3.)P(10,5) = 10!/5! = 10 × 9 × 8 × 7 × 6
4.)P(10,3) = 10!/7! = 10 × 9 × 8
5.)4 × 4 × 4 = 43
6.)P(5,3) = 5!/2! = 5 × 4 × 3
7.)Event (Choose 2 letters) Event (Choose 3 digits)
26 × 26 = 262 arrangements of letters 10 × 10 × 10 = 103 arrangements of digits# Arrangements (Both) = # Ways Choose 2 letters × # Ways choose 3 digits
= 26 × 26 × 10 × 10 × 10 = 262 × 103
8.)Dependencies make some arrangements not possible. First the digit 0 cannot be used in
the 1st position else the result is a 3-digit not 4-digit number. Also, the last digit must a 0
or 2 in order to be an even number.
Group the options for the 1st digit and then count the arrangements of the other 3 digits
by working from the last digit backward.
1: 2 ways to choose last digit(0 or 2) × 2 ways to choose remaining × 1 way for last digit
= 4 ways to start with a 1.
2: 2 ways to choose last digit(0 or 2) × 2 ways to choose remaining × 1 way for last digit
= 4 ways to start with a 2.
3: 2 ways to choose last digit(0 or 2) × 2 ways to choose remaining × 1 way for last digit
= 4 ways to start with a 3.
In total, that is 4 + 4 + 4 = 12 ways to create 4-digit even numbers.
9.)If 3 positions are determined to be together then 3 names are reduced to essentially 1
group which makes a total of 1 group + 3 other names = 4 objects to choose for ordering.
4 unlike objects can be arranged in 4! ways.
10.)There are 5! ways to arrange 5 unlike letters. Among that number of arrangements
exactly half will have i to the left of g and the other half will have i to the right of g. As a
result, 5!/2 arrangements of the letters SAIGE exist with "i" to the right of "g".
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Answers & Comments
Answer:
1.) 5 choices × 5 choices × 5 choices = 53 possible 3 digit codes including repeats
2.)Choose two letters from four = P(4,2) = 4!/2! = 4 choices × 3 choices = 12. Order matters
i.e. DA is a different word than AD.
3.)P(10,5) = 10!/5! = 10 × 9 × 8 × 7 × 6
4.)P(10,3) = 10!/7! = 10 × 9 × 8
5.)4 × 4 × 4 = 43
6.)P(5,3) = 5!/2! = 5 × 4 × 3
7.)Event (Choose 2 letters) Event (Choose 3 digits)
26 × 26 = 262 arrangements of letters 10 × 10 × 10 = 103 arrangements of digits# Arrangements (Both) = # Ways Choose 2 letters × # Ways choose 3 digits
= 26 × 26 × 10 × 10 × 10 = 262 × 103
8.)Dependencies make some arrangements not possible. First the digit 0 cannot be used in
the 1st position else the result is a 3-digit not 4-digit number. Also, the last digit must a 0
or 2 in order to be an even number.
Group the options for the 1st digit and then count the arrangements of the other 3 digits
by working from the last digit backward.
1: 2 ways to choose last digit(0 or 2) × 2 ways to choose remaining × 1 way for last digit
= 4 ways to start with a 1.
2: 2 ways to choose last digit(0 or 2) × 2 ways to choose remaining × 1 way for last digit
= 4 ways to start with a 2.
3: 2 ways to choose last digit(0 or 2) × 2 ways to choose remaining × 1 way for last digit
= 4 ways to start with a 3.
In total, that is 4 + 4 + 4 = 12 ways to create 4-digit even numbers.
9.)If 3 positions are determined to be together then 3 names are reduced to essentially 1
group which makes a total of 1 group + 3 other names = 4 objects to choose for ordering.
4 unlike objects can be arranged in 4! ways.
10.)There are 5! ways to arrange 5 unlike letters. Among that number of arrangements
exactly half will have i to the left of g and the other half will have i to the right of g. As a
result, 5!/2 arrangements of the letters SAIGE exist with "i" to the right of "g".