Math Fall 2011 Exam 2 Solutions - November 1, 2011

Transcription

1 Math Fall 011 Exam Solutions - November 1, 011 NAME: STUDENT ID: This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 5 problems for a total of 100 points.

2 1. 0 points) a) How many integers between 100 and 999 begin with an odd digit? Solution. There are 5 ways to select the first odd digit. Then there are 10 ways to select the second digit and 10 ways to select the third. By the multiplication principle there are = 500 integers. b) How many license plates can be made using either letters followed by 3 digits or 3 letters followed by digits? Solution. First we count the license plates with letters and 3 digits. There are 6 possibilities for the first letter, 6 possibilities for the second letter and 10 possibilities to select each of the three digits. Thus there are = possibilities total. Similarly, for the plates with 3 letters and digits we have = possibilities. By the summation principle there are = different license plates. c) What is the minimum number of students required in a class to be sure that at least six will receive the same grade? Possible grades A, B, C, D, F) Solution. We use pigeonhole principle and get that the maximum number of students among which no six have the same grade is 5 5 = 5. Thus the minimum number of students with at least six having the same grade is 6. d) How many ways are there to arrange n men and n women in a row if men and women alternate? Solution. There are n! ways to arrange the men in a row and then there are n!ways to arrange the women in a raw. we can combine them in two ways, starting with a women or with a ma. Therefore, there are n!) ways total.

3 . 0 points) A candy merchant carries bubble gum balls in seven different colors. In how many ways can you a) Buy ten bubble gum balls? Solution. We need to select 10 objects bubble gum balls) from a set with 7 different elements colors) when repetition is allowed. We get C10+7-1,10)=C16,10) ways to buy ten bubble gum balls. b) Buy ten bubble gum balls, with exactly three red ones? Solution. After picking three red ones we need to select another seven out of the remaining six colors. We get C1,7) ways. c) Buy ten bubble gum balls including at least one of each color? Solution. We pick seven bubble balls each of a different color. Then we can select the remaining three balls in C7+3-1,3)=C9,3) different ways. d) Buy ten bubble gum balls of at most six different colors? Solution. There are C16,10) ways to buy 10 bubble balls and C9,3) ways to have all seven different colors. Therefore there are C16,10)- C9,3) ways to buy ten bubble gum balls of at most six different colors.

4 3. 0 points) a) In how many ways can the letters of the word NUNAVUT be rearranged? Solution. There are N, U and A,V,T. First we place the two N. This can be done in C7,) ways. After that we have five spots left, so we can place U in C5,) ways. Then there are 3! ways to arrange A,V,T on the remaining 3 spots. Thus, there are C7,)C5,)3!=160 ways. b) Find the coefficient of x 6 in 16x 1 x )1. Simplify your answer. Solution. According to the binomial theorem 16x 1 x )1 = 1 j=0 C1,j)16x ) 1 j x) j 1 = C1,j) 1) j 16 1 j x 4 3j j j=0 To find the coefficient of x 6 we set 4 3j = 6 and get j = 10. Thus the coefficient is. C1,10) 1) = 33

5 4. 0 points) In how many ways can five rings be put onto the four fingers of one hand no thumb) if a) the rings are all identical; Solution. We need to distribute 5 indistinguishable objects rings) into 4 distinguishable boxes fingers). According to the formula we have C4+5-1,5)=C8,5)=56 ways to do it. b) the rings are all different. Solution. From part a) there are 56 ways to place rings on the fingers. For each of this placements we can rearrange the rings in 5! ways. Therefore, there are = 670 ways to put five different rings onto the four fingers of one hand.

6 5. 0 points) Solve the recurrence relation a n = 5a n 1 a n + 3n, n, given a 0 = 3, a 1 = 11 Solution. The characteristic equation of the associated homogeneous equation is r 5r + = 0. It has two roots r 1 = 5 17, r = 5+ 17, so the solution 5 to the homogeneous equation is c c 17. We look for the particular solution in the form p = a+bn+cn. We obtain a+bn + cn = 5[a + bn 1) + cn 1) ] [a + bn ) + cn ) ] + 3n = 5[a b + c + b c)n + cn ] [a b + 4c + b 4c)n + cn ] + 3n = 3a b 3c + 3b c)n + 3c + 3)n Setting equal the coefficients near different powers of n we get a system a = 3a b 3c; b = 3b c; c = 3c + 3 with solution a = 3, b = 3, c = 3. Therefore the solution to the inhomogeneous equation is a n = a n = c ) n + c 5 ) n n 3 n 5 The initial conditions give c 1 + c 3 = 3 and c c 17 6 = 11. This gives c 1 = 17 and c = 17. Therefore, a n = ) n ) n 3 3 n 3 n