A mathematics professor wishes to schedule an appointment with each of her 8 tea

Question

A mathematics professor wishes to schedule an appointment with each of her 8 teaching assistants, four men and four women, to discuss her calculus course. Suppose that all possible orderings of appointments are equally likely to be selected.

a.      How many possible orderings of appointments are there if each person is identified by their name?

b.      What is the probability that the ordering alternates between male name and female name (it can start either with male or female as long as it alternates)?

c.      What is the probability, after seeing 6 students, the professor will have seen all the females TAs? HINT: the 7th and 8th appointments must be male names.  

Explanation / Answer

a. Number of possible arrangments = 8! = 40320

b. Number of alternating arrangements with male starting = Arrangements of males at odd places( i.e. 1st, 3rd, 5th and 7th)* Arrangements of females at even places = (4!)*(4!)

Same will be the number of arrangements with female starting.

So, total number of arrangements = 2*4!*4! = 2*24*24 = 1152

Probability = 1152/40320 = 0.0285

c) Probability that last two students are male = Possible arrangments for last two students * other six

= (4C2*2!)*6! = 6*2*6! = 8640

Probability = 8640/40320 = 0.2142

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