math help v. discrete math... test tomorrow!
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math help v. discrete math... test tomorrow!
can someone please help me with the following math questions i have a test tomorrow... thanks!
RECURSIVELY DEFINED SEQUENCES:
ii. Solve an = 4an−1 + 5an−2 + 3n, n >= 2, given a0 = 4, a1 = −1.
PERMUTATIONS:
3. A committee of seven is to be chosen from eight men and nine women.
a. How many such committee contain at leat six women?
b. How many such committees contain either Bob or Alice, but not both?
RECURSIVELY DEFINED SEQUENCES:
ii. Solve an = 4an−1 + 5an−2 + 3n, n >= 2, given a0 = 4, a1 = −1.
PERMUTATIONS:
3. A committee of seven is to be chosen from eight men and nine women.
a. How many such committee contain at leat six women?
b. How many such committees contain either Bob or Alice, but not both?
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RDS:
First, solve the homogeneous equation:
an = 4an-1 + 5an-2
an - 4an-1 - 5an-2 = 0
t^2 - 4t - 5 = 0
(t - 5)(t + 1) = 0
t = 5, -1
an = c1*5^n + c2*(-1)^n
Because t = 1 is not a root of the homogeneous equation, a particular solution of the nonhomogeneous equation takes the form:
c3n + c4
Thus, the general solution of the nonhomogeneous equation is:
an = c1*5^n + c2*(-1)^n + c3n + c4
I leave it to you to solve for c1, c2, c3, and c4.
P:
a. Calculate the number that contain exactly 6 women, the number that contain exactly 7 women, then add the two.
b. The easiest route is probably to compute the number that contain both Bob and Alice, and the number that contain neither Bob nor Alice, then subtract these two numbers from the total possible containing anyone.
First, solve the homogeneous equation:
an = 4an-1 + 5an-2
an - 4an-1 - 5an-2 = 0
t^2 - 4t - 5 = 0
(t - 5)(t + 1) = 0
t = 5, -1
an = c1*5^n + c2*(-1)^n
Because t = 1 is not a root of the homogeneous equation, a particular solution of the nonhomogeneous equation takes the form:
c3n + c4
Thus, the general solution of the nonhomogeneous equation is:
an = c1*5^n + c2*(-1)^n + c3n + c4
I leave it to you to solve for c1, c2, c3, and c4.
P:
a. Calculate the number that contain exactly 6 women, the number that contain exactly 7 women, then add the two.
b. The easiest route is probably to compute the number that contain both Bob and Alice, and the number that contain neither Bob nor Alice, then subtract these two numbers from the total possible containing anyone.
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From: Yorba Linda, CA
Originally Posted by Elistan' date='Mar 25 2007, 07:54 PM
3a - 9 choose 6 times eight choose one = 672
3b - 16 choose 6 (committees with Bob) plus 16 choose 6 (committees with Alice) = 16016
3b - 16 choose 6 (committees with Bob) plus 16 choose 6 (committees with Alice) = 16016
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From: Yorba Linda, CA
Originally Posted by DangerZone9K' date='Mar 25 2007, 08:20 PM
thats not the answer?
Originally Posted by DangerZone9K' date='Mar 25 2007, 08:20 PM
do you know what is?
Furthermore, I wrote (four posts up) exactly how to calculate them.
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From: Longmont, CO
Originally Posted by magician' date='Mar 25 2007, 11:18 PM
Alas, not even close.
Past my bedtime. 
But for 3a - 9 choose 6 would give you the different ways to put the "at least six women" onto the committee. Then you're completely free to choose the remaining person from the non-picked people - 11 different choices. So the answer would be 924, no?
9 choose 6 times 8 gives the number of committees with exactly six women, doesn't it? Then add to that 9 choose 7, the number of committees with exactly six women. Yielding 708...
What am I missing?