Consider the preferences lists of the stable marriage algorithm. assume that the ranking lists of all women by the men are the same, and analogously, the ranking of all men by the women are the same. in other words, there is a consensus between the women on who the favorite man is, who the second favorite man is, and so on. prove that given this scenario, there is only one stable matching. what is it? (note that in general, the tma finds one of possible multiple stable matching)

can you me with homework?

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the answer   is 200

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this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right

i believe the answer is letter c) 8

step-by-step explanation: