(problem 1a) Two 6-sided dice are thrown, one of which is red and the other green. In
how many ways can at least one 4 be rolled?
The number of outcomes with at least one 4 is .
We use complements to enumerate sets.
The complement
Rule for Complements The number of elements in is the number of elements in
minus the number of elements in :
example 1 Two 6-sided dice are thrown, one of which is red and the other green. In
how many ways can at least one 6 be rolled.
We let be the set of all 36 possible outcomes of the form (red, green). The set which we would like to enumerate is the subset of which contains a 6 as a component in at least one position. The complement, , is the set of outcomes which do not contain a 6 in either the red or the green component. It is easier to count the number of elements in than it is to count the number of elements in , so we will use the rule for Complements. since each component can be a number from 1 to 5. Now, the Rule for Complements gives Thus there are 11 outcomes where at least one of the die rolls is a 6.
We let be the set of all 36 possible outcomes of the form (red, green). The set which we would like to enumerate is the subset of which contains a 6 as a component in at least one position. The complement, , is the set of outcomes which do not contain a 6 in either the red or the green component. It is easier to count the number of elements in than it is to count the number of elements in , so we will use the rule for Complements. since each component can be a number from 1 to 5. Now, the Rule for Complements gives Thus there are 11 outcomes where at least one of the die rolls is a 6.
(problem 1b) Two 6-sided dice are thrown, one of which is red and the other green. In
how many ways can their sum be at least 3?
The number of outcomes whose sum is at least 3 is .
The number of outcomes whose sum is at least 3 is .
example 2 Two 6-sided dice are thrown, one of which is red and the other green. In
how many ways can the sum be at most 10?
We let be the set of all 36 possible outcomes of the form (red, green). The set which we would like to enumerate is the subset of whose components sum to at most 10. This means that their sum can be 2, 3, 4, ..., 10. This is most of the possible sums, so most likely, it will be easier to enumerate by examining its complement. contains the outcomes whose sum is more than 10. That is, the sum is either 11 or 12. There are two outcomes that give a sum of 11, namely and and there is one outcome that gives a sum of 12, . Thus, Note that was partitioned into two subsets here. Now, by the Rule for Complements, Thus, there are 33 outcomes which have a sum of at most 10.
We let be the set of all 36 possible outcomes of the form (red, green). The set which we would like to enumerate is the subset of whose components sum to at most 10. This means that their sum can be 2, 3, 4, ..., 10. This is most of the possible sums, so most likely, it will be easier to enumerate by examining its complement. contains the outcomes whose sum is more than 10. That is, the sum is either 11 or 12. There are two outcomes that give a sum of 11, namely and and there is one outcome that gives a sum of 12, . Thus, Note that was partitioned into two subsets here. Now, by the Rule for Complements, Thus, there are 33 outcomes which have a sum of at most 10.
(problem 2) Two 6-sided dice are thrown, one of which is red and the other green. In
how many ways can their sum be at least 5?
The number of outcomes whose sum is at least 5 is .
The number of outcomes whose sum is at least 5 is .
example 3 Computer passwords are eight characters long where a character can be
either an upper case letter, lower case letter, digit 0-9, or one of 8 special symbols.
How many passwords are possible if a password must contain at least one special
symbol?
With no restrictions on the symbols, we have seen that the total number of possibilities is To enumerate the passwords with at least one special symbol, it will be helpful to consider the complement. That is, passwords with no special symbols. If we use only capital letters, lower case letters and digits to construct our password, then there are possibilities. Thus, the Rule for Complements tells us that the number of passwords which contain at least one special symbol is Thus, there are over 358 trillion possible passwords if we insist on at least one special symbol.
With no restrictions on the symbols, we have seen that the total number of possibilities is To enumerate the passwords with at least one special symbol, it will be helpful to consider the complement. That is, passwords with no special symbols. If we use only capital letters, lower case letters and digits to construct our password, then there are possibilities. Thus, the Rule for Complements tells us that the number of passwords which contain at least one special symbol is Thus, there are over 358 trillion possible passwords if we insist on at least one special symbol.
(problem 3a) Computer passwords are eight characters long where a character can be
either an upper case letter, lower case letter, digit 0-9, or one of 8 special symbols.
How many passwords are possible if a password must contain at least one capital
letter?
The number of passwords with at least one capital letter is .
The number of passwords with at least one capital letter is .
example 4 How many bit matrices contain at least one 1?
If we let be the set of all bit matrices, then by the FPC, . We wish to find , where is the subset of consisting of bit matrices with at least one 1. The complement, is the set of bit matrices containing zero 1’s and it is easy to enumerate. In fact, , since the matrix must contain all 0’s. Thus, and there are fifteen bit matrices containing at least one 1.
If we let be the set of all bit matrices, then by the FPC, . We wish to find , where is the subset of consisting of bit matrices with at least one 1. The complement, is the set of bit matrices containing zero 1’s and it is easy to enumerate. In fact, , since the matrix must contain all 0’s. Thus, and there are fifteen bit matrices containing at least one 1.
example 5 A coin is flipped 5 times and the sequence of heads and tails is recorded.
How many possible outcomes contain at least 2 heads?
By the FPC, the total number of outcomes for the 5 coin flips is . The condition “at least two heads” contains the possibilities of 2, 3, 4 or 5 heads. On the other hand, the complement consists of only the outcomes 0 or 1 heads, suggesting that we consider using the Rule for Complements. Since there is 1 way to get 0 heads (namely TTTTT), and there are 5 ways to get exactly 1 heads (the H can go in any one of 5 places, with the other 4 places being T’s), we have By the Rule for Complements, Thus, there are 26 outcomes consisting of 2 or more heads in 5 coin flips.
By the FPC, the total number of outcomes for the 5 coin flips is . The condition “at least two heads” contains the possibilities of 2, 3, 4 or 5 heads. On the other hand, the complement consists of only the outcomes 0 or 1 heads, suggesting that we consider using the Rule for Complements. Since there is 1 way to get 0 heads (namely TTTTT), and there are 5 ways to get exactly 1 heads (the H can go in any one of 5 places, with the other 4 places being T’s), we have By the Rule for Complements, Thus, there are 26 outcomes consisting of 2 or more heads in 5 coin flips.