Exam 1 Questions with Solution - First-Year Interest Group Seminar | F A 1, Exams of Art

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Examination 1 Solutions CS 336

1. [5] Given sets A and B , each of cardinality n , how many functions map in a one- to-one fashion onto

≥ 1 A

B?

Let A = { a 1 (^) , a 2 ,..., an }and f : A  onto 1 1−→ B. There are n options for the value of f ( a 1 (^) ) and, given that, n − 1 options for the value of f ( a 2 ), …., and one option for the value of f ( an ). Since B also has cardinality this function is automatically onto. Thus, there are such one-to-one and onto functions.

n n!

2 a. [5] Given the set of symbols { , , how many different strings of length exist (allowing repetitions)?

r a (^) 1 a (^) 2 ,..., a (^) r } n ≥ 1

For each of n positions there are r options. Thus, there are rn such strings.

b. [10] Given the set of r symbols , how many different strings of length

exist that contain at least one and at least one? (Assume .)

{ a (^) 1 , a (^) 2 ,..., a (^) r } n ≥ 2 a 1 a 2 r ≥ 2

There are ( strings that avoid the element a and the same number that avoid

. There are subsets that avoid both elements so there are strings that either avoid a or avoid. Considering the complement, there are r r strings that contain at least one a and at least one a.

r − 1) n ( r − n (^) ( r −2)

2

1

n

a 2 2(

2. n r − 1) − n 1 −2)

a 2 n (^) − 2( − 1) n (^) + ( r 1

3. [10] Present a combinatorial argument that for all positive integers n :

0

n n k k

n = k

 

=  

 

∑.

Consider as a model strings of length using the characters from the set { ,. For each positions there are 3 options so there are 3 such strings. Alternatively, let k represent the number of positions in the string not occupied by (i.e., thus, occupied by either b or ). The value of k can vary between 0 and. For a fixed

number of b s and s, there are

n

n k

 



 

a b c , }

k

n

k

n a c n

c n k

 

 

 

c (^)  ways to determine the positions to be

occupied by the s and s and then choices (either b or c ) for each of these

positions, for a total of possibilities. The remaining

b

2 k^ n − k positions must be

occupied by s. Summing over all possible values of. We have such

strings and this must equal.

a k

≥

0

n k k

n = k

 

 

 

∑

3 n

 

 =

 

∪ B ∪

n

,

n 3

^ n    

,

,

a 1 (^) , a 2

,...,

n r n

+

b. [10] Present a combinatorial argument that for all integers n 3 : (^3 3) 3 2 3 3 3 2

n n n n n

   

  +^ ⋅^  +

   

( Hint: Consider three pairwise disjoint sets of cardinality n .)

Let and C be pairwise disjoint sets of cardinality. Consider as a model the number of subsets of of cardinality 3. Since the cardinality of

is 3 , there are



 such subsets of cardinality 3. Now consider that

all three elements could come from the same set or C , that two could come from one and one comes from another, and that each of the three could come from

a different set. In the first case, there are 3 options for the set and then

A B , ,

∪ B ∪

A C

n  

A C n





A B ,

n n n

 

ways of

selecting the subset. In the second case, there are there are 3 options for the set

from which 2 elements are selected, then ways of selecting those two elements,

and 2 choices for the set from which only one element is selected, and finally options for that selection. In the final case, there are possible selections from

each of the three sets. The total is 3 3

^ n      n

n

 

+ ⋅  

 

 ^ +

 

and this must equal

 

 .

n  



4. [10] A multiset is similar to a set in that order is irrelevant but multiple copies of elements are allowed. For example, the sets {1 and { are identical and each has

cardinality three but the multisets {1 and { are different and the first has cardinality three but the second has cardinality six. How many multisets of cardinality are there that employ elements from?

2,3}

2,3}

,...,

1,1,1, 2, 2,3}

1,1,1, 2, 2,3}

n a r

Let us label r bins and consider the number of ways of placing n indistinguishable balls into the bins. The placements of balls into bins is in one-to- one correspondence with multisets of cardinality n that employ elements from

. There are

a 1 (^) , a 2 ar r

a 1 (^) , a (^) 2 ,..., ar

 −

 such placements of balls in bins so there is the

same number of multisets.





 

5 a. [5] How many strings are there of length using elements from the set {

if repetition is not allowed. (Assume

k ≥ 1 1, 2,..., } n k ≤ n ).

Since there are options for the first element of the string, options for the second element of the strings, …, and

n n − 1 n − ( k − 1 ) options for the last element there

are

!

( 1) ( ( 1))

( )

n n n n k n k

⋅ − ⋅ ⋅ − − =

−

"

!

such strings.

b. [10] Now assume each of the different strings in part a. is equally likely. What is the probability that the minimum of the k elements is less than or equal to r , where 1 ≤ r ≤ n − k + 1?

Now we must count how many of these strings have minimum of the k elements is less than or equal to r. Alternatively we could count how many of these strings have minimum of the k elements is strictly greater than r. If the minimum of the k elements is strictly greater than r , then there are only n − r choices for first element of the string, options for the second element of the strings, …, and options for the last element. So there are

n − r − 1 n − r − ( k − 1) ( ( ( 1)) ( )

n r n r k n k

)!

!

( n r ) ( n r 1)

−

− ⋅ − − ⋅"⋅ − − − =

−

such strings. The probability of

such a string is

(

!

n r n

− )!

and the probability that the minimum of the k elements is

less than or equal to r is

( )!

!

n r n

−

−.

6. a. [5] a. How many permutations of a, b, c, d, and e have both a to the left of b and b to the left of c?

There are 5 positions in the string for the d and then 4 positions for the e. Once these are fixed, the positions for a , b , and c are determined since a must be to the left of b and b to the left of c and these must fill the three remaining positions. Thus there are 5 4⋅ = 20 such permutations.

b. [10] Assume all such permutations are equally likely, what is the probability that the permutation begins with a given that it has both a to the left of b and b to the left of c?

If the permutation begins with a, then there are only 4 positions in the string for the d and then 3 positions for the e. Thus there are 12 such strings with a in the first position and b to the left of c (a will automatically be to the left of b). The probability that the permutation begins with a given that it has both a to the left of b and b to the left of c is 12 / 20.