Exam 1 Math 375 Spring Instructions: Write one answer per blank sheet of paper.

Transcription

1 Exam 1 Math 375 Spring 2014 Name: Instructions: Write one answer per blan sheet of paper. 1. The Math Club consists of three freshmen, four sophomores, and six juniors. A team of five people is selected from the club for a math competition. The team must include at least one junior. How many teams can be created? Only membership matters, not order. Solution: There are ( ) 13 5 ways to choose a team without restriction. Unwanted teams are those that exclude juniors, and there are ( 7 5) of these. So there are ( ) ( ) = 1266 possible teams. Discussion: Why 6 ( ) 12 4 = 2970 is incorrect: If you choose a junior in 6 ways then choose 4 more people in ( ) 12 4 ways then there are two ways to produce the team {F 1, F 2, F 3, J 1, J 2 }: You could pic J 1 first, and then round out the team with {F 1, F 2, F 3, J 2 }, or you could pic J 2 first, then round out the team with {F 1, F 2, F 3, J 1 }. In general, you would count a team containing juniors in it different times. 1

2 2. Four types of pizza are served at the Math Club: cheese, veggie, sausage, and pepperoni. You are very hungry. How many ways are there for you to tae ten slices of pizza assuming that at least one slice of each type is selected? Assume only quantities matter. Solution: After you pic a slice of each, you need to pic 6 more slices without restriction. The number of solutions to x c + x v + x s + x p = 6 is equal to the number of rearrangements of 6 dots and 3 bars, which is 9! 6!3! = 84. Discussion: Why x c + x v + x s + x p = 10 is incorrect: you could end with all cheese slices. Why binomial coefficients are incorrect: this problems deals with multisets, not subsets. Why 4 10 = is incorrect: this is the number of pizza sequences you could form, which violates the condition that only quantities (how many) matter. Why 4! = 24 is incorrect: this is the number of ways you can rearrange a cheese slice, a veggie slice, a sausage slice, and a pepperoni slice on your plate, which again violates the condition that only quantities matter. 2

3 3. How many was are there to rearrange the letters in the word PEPPERONI in such a way that at least two letters are between every pair of Ps? For example, NPEEPORIP. Solution: Choose the P spacing, then choose the order of the other letters. P P P : We are counting solutions to x 1 + x 2 + x 3 + x 4 = 6 where x 1 0, x 2 2, x 3 2, x 4 0. Rescaling, we are counting solutions to 5! y 1 + y 2 + y 3 + y 4 = 2. 2 dots, 3 bars,. 2!3! E 2 R 1 O 1 N 1 I 1 6! :. 2!1!1!1!1! 5! 6! Total: = !3! 2!1!1!1!1! Discussion: It might be possible to use an ALL BAD approach, but 9! only if you can account for every bad possibility. This yields BAD. 3!2! Unfortunately, there are too many BAD cases to efficiently count in a short amount of time, which should tell you to move on to a more efficient method. 3

4 4. Simplify: =1 ( ) ( 2) = Solution: Answer: 200. (x + y) n = n =0 ( ) n x y n (x + 1) = x =0 (x + 1) 99 1 = x 1 =0 x(x + 1) 99 = x =0 ( 2)( 1) 99 = ( 2) Discussion: If you use y = 0 then y n becomes 0 for < n and you are not producing all the terms in the sum. Writing out all the terms, the expression is ( ) 1 1( 2) 1 + ( ) 2 2( 2) 2 + ( ) 3 3( 2) 3 +. You cannot factor out ( 2) because varies and there is no term common to all expressions that can be factored out (except perhaps 2). =0 4

5 5. Find a collection of objects to count and identify two different methods of counting them so that, using the first method, you get n =0 ( ) n (2 + 1), and using the second method, you get n2 n 2, thereby proving ( ) ( ) ( ) n n n = n2 n Solution: Count subsets of [n] of odd size that contain a circled number. Method 1: choose a number to circle, then choose an even-sized subset of the remaining n 1 numbers to accompany the circled number. This yields n2 n 2. Method 2: the number of these of size is found by choosing numbers from [n], then choosing one of these numbers to circle. This yields ( n 2+1) (2+1), and summing over all odd sizes we get the summation formula. Discussion: If you count the odd elements in [n] you are going to get either n/2 or (n + 1)/2, depending on whether n is even or odd. If you differentiate the binomial theorem formula you will obtain all possible ( n ) contributions, not just the (2 + 1) ( n 2+1) terms. 5