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Exam 1 with Solution for First Year Interest Group Seminar | N 1, Exams of Health sciences

Material Type: Exam; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/30/2009

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koofers-user-n6h 🇺🇸

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Examination 1 Solutions CS 336

**1. The important issue is the logic you used to arrive at your answer.

  1. Use extra paper to determine your solutions then neatly transcribe them onto these sheets.
  2. Do not submit the scratch sheets. However, all of the logic necessary to obtain the solution should be on these sheets.
  3. Comment on all logical flaws and omissions and enclose the** **comments in boxes
  4. [5]** Suppose all rolls of a six-side die are equally likely. What is the probability the roll is a

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 ≤ kn

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 ,

 −^3

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