Exam 9: Counting and Probability

Correct Answer:

Verified

First note that there is one string of length zero that does not contain the pattern bb, namely
ϵ, the null string; there are two strings of length one that do not contain the pattern bb, namely
a and b; and there are three strings of length three that do not contain the pattern bb, namely
aa, ab, and ba. Thus the initial conditions for the recurrence relation are $$a _ { 0 } = 1 , a _ { 1 } = 2 , \text { and } a _ { 2 } = 3$$ Let $k$ be any integer with $k \geq 2$ . Any string of length $k$ that does not contain the pattern $b b$ starts either with an $a$ or with a $b$ . If it starts with an $a$ , this can be followed by any string of $k - 1 b$ 's that does not contain the pattern $b b$ . There are $a _ { k - 1 }$ of these. If the string starts with a $b$ , then it must be followed by an $a$ and that can be followed by any string of length $k - 2$ that does not contain the pattern $b b$ . There are $a _ { k - 2 }$ of these. Therefore, for all integers $k \geq 2$ ,
$$a _ { k } = a _ { k - 1 } + a _ { k - 2 }$$

Correct Answer:

Verified

Yes.
Solution 1: There are 25 possible remainders that can be obtained when an integer is divided by 12, namely all the integers from 0 through 24. Apply the pigeonhole principle, thinking of the elements of the set of 30 integers as the pigeons and the possible remainders as the pigeonholes. Each pigeon flies into the pigeonhole that is the remainder obtained when it is divided by 25. Since 30 > 25, the pigeonhole principle says that at least two pigeons must fly into the same pigeonhole. So at least two of the numbers must have the same remainder when divided by 25.
Solution 2: Let $X$ be the set of 30 integers and $Y$ the set of all possible remainders obtained through division by 30 , and consider the function $R$ from $X$ (the pigeons) to $Y$ (the pigeonholes) defined by the rule: $R ( n ) = n \bmod 25$ (= the remainder obtained by the integer division of $n$ by 25). Now $X$ has 30 elements and $Y$ has 25 elements (the integers from 0 through 24). Hence by the pigeonhole principle, $R$ is not one-to-one: $R \left( n _ { 1 } \right) = R \left( n _ { 2 } \right)$ for some integers $n _ { 1 }$ and $n _ { 2 }$ with $n _ { 1 } \neq n _ { 2 }$ . But this means that $n _ { 1 }$ and $n _ { 2 }$ have the same remainder when divided by 25 .

Correct Answer:

Verified

Yes.
Solution 1: There are 12 possible remainders that can be obtained when an integer is divided
by 12, namely 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Apply the pigeonhole principle, thinking of the elements of the set of 15 integers as the pigeons and the possible remainders as the pigeonholes. Each pigeon flies into the pigeonhole that is the remainder obtained when it is divided by 12. Since 15 > 12, the pigeonhole principle says that at least two pigeons must fly into the same pigeonhole. So at least two of the numbers must have the same remainder when divided by 12.
Solution 2: Let $X$ be the set of fifteen integers and $Y$ the set of all possible remainders obtained through division by 12, and consider the function $R$ from $X$ (the pigeons) to $Y$ (the pigeonholes) defined by the rule: $R ( n ) = n \bmod 12 ( =$ the remainder obtained by the integer division of $n$ by 12). Now $X$ has 15 elements and $Y$ has 12 elements $( 0,1,2,3,4,5,6,7,8$ , $9,10,11 )$ . Hence by the pigeonhole principle, $R$ is not one-to-one: $R \left( n _ { 1 } \right) = R \left( n _ { 2 } \right)$ for some integers $n _ { 1 }$ and $n _ { 2 }$ with $n _ { 1 } \neq n _ { 2 }$ . But this means that $n _ { 1 }$ and $n _ { 2 }$ have the same remainder when divided by 12 .