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Discrete math question regarding combination equations?

Discrete math question regarding combination equations? Topic: C case switch example in c
July 18, 2019 / By Kristina
Question: so there are two equations for combinations with repetitions: one is: C(n+r-1,r) and the other is: C(n+r-1,n-1) ive seen separate examples using both but they can never be overlapped/switched in those cases, when do i use which equation then? which one is universal? how can i tell when i should use them? thanks for your time.
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Jeanie Jeanie | 1 day ago
There are 12 techniques of sharing the 1st motor vehicle. as quickly as this is finished there are 11 techniques of sharing the 2nd. So the two first automobiles provide 12X11 techniques of being shared . There are 10 techniques of sharing the third motor vehicle and 9 techniques of sharing the 4th motor vehicle the finished style of techniques is of sharing the 4 automobiles is 12X11X10X9 :-)
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Jeanie Originally Answered: Discrete Math, how does implication make sense?
In your example, the conclusion of your implication is only dependent on the one condition set out by the hypothesis, that is it sunny today. The way that I think of it is as such: Say you promise someone that if it is sunny then you will go to beach. The only way that someone could rightfully accuse you of breaking this promise is if it was sunny and you did not go. The statement says nothing of what you will do when it is not sunny and hence cannot be false when referring to anything other than a sunny day.
Jeanie Originally Answered: Discrete Math, how does implication make sense?
p>q implies: not q>not p, that's it if it's sunny, I'll go to the beach If I didn't go to the beach, it wasn't sunny

Jeanie Originally Answered: Need help with a Discrete Structures math proof? Simple, but need help getting started?
Have you learned modular arithmetic? If so, Suppose a = 0 (mod 3). Then a^3 - a = 0 - 0 = 0 (mod 3). Suppose a = 1 (mod 3). Then a^3 - a = 1 - 1 = 0 (mod 3). Suppose a = 2 (mod 3). Then a^3 - a = 2 - 2 = 0 (mod 3). Hence, 3|(a^3 - a) for ∀a ∈ Z. If you haven't learned modular arithmetic: Let a ∈ Z. •Suppose a = 3k, k ∈ Z. a^3 - a = (3k)^3 - 3k = 27k^3 - 3k = 3(9k^3 - k). So 3|(a^3 - a). •Suppose a = 3k + 1, k ∈ Z. a^3 - a = (3k + 1)^3 - (3k + 1) = 27k^3 + 27k^2 + 9k + 1 - 3k - 1 = 27k^3 + 27k^2 + 6k = 3(9k^3 + 9k^2 + 2k). So 3|(a^3 - a). •Suppose a = 3k + 2, k ∈ Z. a^3 - a = (3k + 2)^3 - (3k + 2) = 27k^3 + 54k^2 + 36k + 8 - 3k - 2 = 27k^3 + 54k^2 + 33k + 6 = 3(9k^3 + 18k^2 + 11k + 2). So 3|(a^3 - a). •For ∀a ∈ Z, a ∈ {3k,3k + 1,3k + 2 : k ∈ Z}, by the division algorithm. Hence, as all cases have been tested, it follows that 3|(a^3 - a) for ∀a ∈ Z.
Jeanie Originally Answered: Need help with a Discrete Structures math proof? Simple, but need help getting started?
Are you aware of modular mathematics? If so, proving this can be a breeze. The modular mathematics approach: We desire to exhibit a^three - a is congruent to zero mod(three) for a=zero, zero-zero is congruent to zero mod(three) for a=a million, a million-a million=zero is congruent to zero mod(three) for a=two, eight-two=6 is congruent to zero mod(three) Thus, a^three -a is congruent to zero mod(three) <===> three divides a^three -a . And we are performed :) But if no longer, we will be able to do it the great distance: three divides a^three - a way 3k=a^three -a, in which okay is an integer [that is what we desire to exhibit] So we've 2 instances: --Case a million: a is even: a=3s, in which s is an excellent integer then, a^three -a = (3s)^three - 3s =27s^three-3s =three[9s^three-s] Since 9s^three -s is an integer, we've the kind 3k --Case two: a is abnormal: a=3s+a million, in which s is an excellent integer then a^three-a = (3s+a million)^three - (3s+a million) =(9s^two+6s+a million)^two(3s+a million) - (3s+a million) =27s^three+9s^two+18s^two+6s+3s+a million - (3s+a million) Combine like phrases and we get: 27s^three+27s^two+6s =three[9s^three+9s^two+2s] Since [9s^three+9s^two+2s] is an integer, we've the kind 3k And we are performed :)

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