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Material Type: Exam; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Unknown 1989;
Typology: Exams
1 / 4
Examination 1 Solutions CS 336
**1. The important issue is the logic you used to arrive at your answer.
six given that it is not one?
The probability of not rolling a one is 5/6. The probability of rolling a six and not rolling a one is 1/6. Thus, the probability of rolling a six given the roll is not a one is
1 / 5 5 / 6
2. a. [10] Present a combinatorial argument that for all n ≥ 1 :
2 1 1
=
n
n
k k
n
( Note : The summation begins with k = 1 .)
Consider the cardinality of the set of non-empty subsets of a set A of n elements. For each element of A, there are two options: either be present in a subset or not. Thus there are 2 total subsets but one of these is empty so there are 2 non- empty subsets of A. Alternatively, let k indicate the cardinality of the subset. Since we are counting non-empty subsets, k ranges from 1 to n. For a fixed value of k ,
there are ways of selecting the k subset elements from the n total elements of A.
Adding this to include all possible cases of k, we obtain and this must equal
n − 1 n
k
n
=
n
k k
n
1 2 n − 1
b. [10] Present a combinatorial argument that for all integers k and n satisfying 3 ≤ k ≤ n
k
n k
n k
n k
n k
n
( Hint: Consider three special elements.)
Consider the number of subsets of size k of a set B of cardinality n. Since n , we may select three elements b of B and let C = B. Thus C has
cardinality n-3 and B = C ∪. We know there are such subsets.
Alternatively, to select k elements of B for a subset there are four options: all k come from C, k -1 come from C and the k th is either b or , k -2come from C and the k-1st and k th are exactly two of b or , or k -3 come from C and all of
and b are present. For the first option, there are (^)
possibilities since
all k come from C. For the second option, there are 3 possibilities, since k -
elements are selected from C and one from the three of or b. For the third
option, there are possibilities, since k -2 elements are selected from C and
one from the three of or is not selected. Lastly, if k -3 come from C and
all of b and b are present, then there are (^)
options. The total is
and this must equal (^)
1 ,^ b 2 , b 3
{ b 1 , b 2 ,
b 2 , b 3
k
n
~ { b 1 , b 2 , b 3 }
k
n
b 3
k
n 3
k
n
b 1 , b 2 ,
k
n
b 3 }
1 ,^ b 2 , b 3
k
n
b 2 ,
b 1 , b 2 ,
k
n
3
k
n
3
k
n
b 1 ,
3
k
n
1 ,^ b 2
3. [10] How many distinct permutations are there of the letters in “mississippi”?
There must be a total of 11 positions in any such permutation. We may select the
four positions to hold the s’s in ways, four positions from the remaining seven
to hold the i’s in ways, two positions from the remaining three to hold the p’s in
ways, and the m must be in the remaining position. Thus, there are
ways of permuting the letters. Written as a multinomial, this is
7. [10] For , how many strings of length n employing the characters { a,b,c } have at least one a?
n ≥ 1
There are 3 strings of length n employing three characters and strings of length n employing just the two characters {b, c}. Therefore, there are 3 strings that have at least one a.
n (^) 2 n n (^) − 2 n