#$&* course Mth 152 06/23 11:30 004. `query 4
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Given Solution:: ** C(9, 6) occurs in the n=9 row, the r=6 position of Pascal’s Triangle, which is the 10th row and the 7th number in the row. C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. (Note that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left.) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK question 11.4.18 clueless check of four of nine possible classrooms. How many of the possible selections will fail to locate the classroom? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:There is are C(9,4) combinations for 4 rooms to be checked. The posssibility for the rooms to have been failed to locate are the C(8,4). C(9, 4) = 9! / ( 4! * (9-4)! )= 128 combinations of the four classrooms to be checked. However, you need the amount that failed to located the classroom. 8! 40320 / = / =70. 4!(4!) 576 Therefore, these 70 fail to include the correct classroom. 128-70=56 total combinations that will be the correct classroom. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** There are C(9,4) possible combinations of the four classrooms to be checked. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check of 4 rooms to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. So there are 126 possible combinations of the four classrooms to be checked, and 70 of these combinations fail to include the ‘right’ classroom. Thus only 126 - 70 = 56 of the possible combinations will include the ‘right’ classroom. So the chance of finding the right classroom is 56 / 126, a little less than 50%. STUDENT QUESTION Can you work out how C (9,4) = 126 and how C (8,4) = 70??? Where did I go wrong??? I worked them out the way I thought I should. Am I getting combinations confused with permutations, or do I just have it totally wrong??? INSTRUCTOR RESPONSE I think you have mixed up permutations and combinations. You calculuated 9! / (9-4)!.. To clarify: C(9, 4) = 9! / ( 4! * (9-4)! ), not 9! / (9-4)!. 9! / (9-4)! = 9 * 8 * 7 * 6, which is the number of possible ordered choices of 4 objects chosen from among 9. This is the permutation. To get the number of unordered choices, you would divide the number of ordered choices by 4! (since every set of 4 choices could have appeared in any of 4! orders). So instead of 9! / (9-4)!., you would divide this by 4! to get 9! / ( 4! (9-4)! ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK question 10.4.33 (previously 11.4.31 /& 30) What sequence by totaling diagonals of Pascal's Triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution:Omitted confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** The totals are 1 1 1 + 1 = 2 1 + 2 = 3 1 + 3 + 1 = 5 Etc. It is easy to see that the numbers are 1, 1, 2, 3, 5, 8, 13, 21, ... and that these numbers form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The rule for the Fibonacci sequence is that the first two numbers are both 1. Then each subsequent number is the sum of the two numbers that precede it.** Regrettably, subsequent problems in this section have been omitted from Section 11.4, as of the 11th edition of the text. The remainder of this Query may therefore be omitted. Query 11.4.42 xxxxxxxxxx (x+y)^8 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. ** Query Add comments on any surprises or insights you experienced as a result of this assignment. ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 xxxxxxxxx The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!