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Discrete Mathematics Questions and Answers
Discrete Mathematics Question Bank with Answer:4. Find the number of combinations of the letter of the word “KEYBOARD” when 5 letter is taken at a time.
Solution: The word “KEYBOARD” has 8 different letters, when 5 letter are taken at a time
The number of combination = ^{n}c_{r} = $\frac{n!}{\left ( nr \right )!r!}$
here n = 8 r = 5 ^{n}C_{r} = $\frac{8!}{\left ( 84 \right )!4!}$
= $\frac{8!}{(4)!4!}=\frac{8*7*6*5*4*3*2*1}{(4*3*2*1)4*3*2*1}$ = 70
5. Find the value of n when the first term of the series is 10, their common difference is 3 and the nth term T_{n} = 64
Solution: Given the first term a = 10, d = 3, T_{n} = 64
Solution: Given a = 2, n = 10
since r is greater than 1 we have sum of series of term formula S_{n} = $\frac{a\left ( r^{n}1 \right )}{\left ( r1 \right )}$
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This chip is essentially a complex array of patterns of propositional logic (p and q, p or q, p implies q, not p, etc.) etched into silicon . see also Data Visualization; Decision Support Systems; Interactive Systems; Physics. Patricia S. Wehman Devlin, Keith. Mathematics: The Science of Patterns. New York: Scientific American Library, 1997. Sangalli, Arturo. The Importance of Being Fuzzy and Other Insights from the Border between Math and Computers. Princeton, NJ: Princeton University Press, 1998.
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Answers:p <=> q ( p q ) ( q p ) p <=> q p q q p p <=> q p q
Answers:The only feasible way for one to occupy a desk without being "seated" is to stand on the desk. So in the first class session, Group A (consisting of 6 students) sits in their respective chairs and so does Group B (6 students). In session 2 Group A stands in their same desk instead of sitting ang Group B switches seats with one another, like musical chairs. Since this arrangement only involves the movement of Group B, there are 216 (6 students times 6 chairs times 6 chairs) possible seating arrangments for Group B for every 1 seating arrangement for Group A. There are 36 possible seating arrangements for Group A (6 students times 6 chairs). So we have 216*36=7776 So Ms Pezzulo has a whopping 7776 seating arrangment options for our single sit down/standup solution. If you want to be technical with it, you should multilply this solution by thenumber of possible bodily positions one could use to occupy a desk without actually putting one's bottom in the seat. LOL. Have at it! I'm going to bed. X) ZZZZZZZZZ
Answers:The key is noticing that: a^3 a = a(a^2 1) = = a(a 1)(a + 1) But a, a 1 and a + 1 are three consecutive integers, so at least one is even, which means that 2(a^3 a). Now, if you divide these three integers by 3, you'll see that at least one of the reaminders must be 0, which is to say that the corresponding integer is divisable by 3, so 3(a^3 a). But gcd(2,3) = 1 so, as a^3 a is divisible by both 2 and 3, it must be divisible by their product, 6.
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