中五數學~排列與組合(4題20分)

1)
(Firstly, accommodate A and B in 2 different floors out of the 3 floors (P(3,2).)
(Then, Accommodate C and D, and either of them in any 1 floor out of the 3floors (3²).)

No. of ways
= P(3,2) x 3²
= (3!/1!) x 9
= 54


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2)
(Choose 3 people (C(6,3)) to take 3 seats of the first row P(4,3).)
(Then, the rest 3 people take 3 seats of the second row P(3,3).)

No. of ways
= C(6,3) x P(4,3) x P(3,3)
= (6!/3!3!) x (4!/1!) x 3!
= 2880


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3)
(Choose 5 children from the 10 children to form the 1st group (C(10,5).)
(Then the rest 5 children form the 2nd group C(5,5).)
(Repetition occurs as the case to choose 5 specific children is identical tothat to choose the rest 5 children (divided by 2)).

No. of ways
= C(10,5) x C(5,5) / 2
= (10!/5!5!) x (5!/5!0!) / 2
= 126


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4)
(Consider only the 8 students other than Mary and Peter.)
(Choose 3 students from the 8 students to form the first group (C(8,3).)
(Then the rest 5 students form the second group (C(5,5).)
(Finally, assign Mary and Peter to these two groups P(2,2).)

No. of ways
= C(8,3) x C(5,5) x P(2,2)
= (8!/3!5!) x (5!/5!0!) x 2!
= 112

1) Since A and B cannot be in the same floor, B can only stay on the other 2 floors.

Thus, 3C1*2C1*3C1*3C1=54//



2)Since in the first row, there are 4 seats in total but only 3 of them are being sat, the no. of combination of seating of the 1st row=7C3*1*4P4=840

The no. of combination of seating of the 2ndt row=7C3*3P3=210

Hence, the number of ways of seating=840+210=1050//



3)10C5(2P2)=504//



4) 3P3*5P5(2P2)=1440//