Four couples go to a theatre and sit side-by-side in one row. How many ways can

Question

Four couples go to a theatre and sit side-by-side in one row. How many ways can they be seated so that at least one couple is not seated next to each other? In congress a five-member committee is to be selected from 8 republicans and 10 democrats. Find the probability that the committee contains at least three republicans How many cards must you pick from a standard deck of 52 cards to be sure of getting at least one face card flack, queen, king)? Explain. Find the number of 5-card poker hands consisting of three of one kind and two of another called a full house, (i.e. 6s, 6d, 6c, ks, kd) For the equation x_1 + x_2 + x_3 + x_4 = 50, how many non-negative integer solutions are there? In a crowd of 3000 people, must at least 8 have the same birthday? Prove that for all nonnegative integers n and r with r

Explanation / Answer

(1)

For four couples there are 8 places to sit. 8 people can be seated in 8! ways.

Consider an event A that all couple are together. this can be done in 4!*(2!)^4

P(A) = 4!*(2!)^4/8!

Required probability that at least one couples are not together = 1 - 4!*(2!)^4/8! = 0.9905

(2)

There are total 8 republican and 10 democrat i.e. 18 members available for selection.

Selecting a committee of 5 from 18 members can be done in 18C5 ways

Selecting at least 3 democrats in committee can be done in 10C3*8C2 + 10C4*8C1 + 10C5*8C0 = 5292

Hence probability of selecting at least 3 democrats in committee = 5292/18C5 = 0.6176

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