F4-5 Math questions

[匿名]

2013-02-14 8:03 pm

Please to slove the following questions

1. A debate team of 7 students are selected from 5 boys and 8 girls.
How many different teams can be formed if
(i) 4 boys are included?
(ii) there are more boys than girls in the team?

2. A 5-member committee is selected from 4 Mathematics teachers and
5 English teachers. In how many ways can the committee be formed if
(i) the committee must include 3 Mathematics teachers?
(ii) the committee must include at least one Mathematics teacher?

3. 3 eggs are chosen from a dozen of eggs, of which 5 are rotten. In how many
ways can the eggs be chosen if at least one good egg is included?

4. 5 cards are taken from a deck of 52 playing cards. Find the number of
combinations that two different pairs are formed

5. There are 10 distinct points on a coordinate plane. If no three points are
collinear, find the number of lines that can be drawn by joining these points.

6.In a hotel, a manager wants to arrange 4 guests to 4 rooms. Two of them
are a couple. Find the number of possible arrangements if
(i) the 4 guests must have their own rooms
(ii) the couple shares the same room and the others have their own rooms

回答 (1)

賣女孩的火柴

2013-02-16 7:27 am

✔ 最佳答案

1.
(a)
B : boys
G : girls

Out of the 5B, 4B are selected (5C4).
Put of the 8G, 3G are selected (8C3).

Number of teams
= 5C4 x 8C3
= (5!/4!1!) x (8!/3!5!)
= 5 x 56
= 280

(b)
No. of teams
= (No. of teams including 5B 2G) + (No. of terms including 4B 3G)
= 5C5 x 8C2 + 280
= (5!/5!0!) x (8!/2!6!) + 280
= 1 x 28 + 280
= 308

2.
(i)
M : Mathematics teacher
E : English teacher

No. of ways that the committee includes 3M and 2E
= 4C3 x 5C2
= (4!/3!1!) x (5!/2!3!)
= 4 x 10
= 40

(ii)
No of ways without restriction
= 9C5
= 9!/5!4!
= 126

No. of ways that the committee includes no M
= No. of ways that the committee includes 5E
= 5C5
= 5!/5!0!
= 1

The required no. of ways
= 126 - 1
= 125

3.
No. of ways withoutrestrictions
= 12C3
= 12!/3!9!
= 220

No. of ways that no good egg is chosen
= No. of ways that 3 rotten eggs are chosen
= 5C3
= 5!/3!2!
= 10

The required no. of ways
= 220 - 10
= 210

4.
(1) Out of the 13 values, choose 2 values (13C2).
(2) For each value, choose 2 cards out of the 4 suits ((4C2)²).
(3) Out of the 44 cards of the rest 11 values, choose 1 card (44C1).

The required no. of combinations
= 13C2 x (4C2)² x (44C1)
= (13!/2!11!) x (4!/2!2!)² x (44!/1!43!)
= 78 x 36 x 44
= 123552

5.
Join any two points to form a line.

No. of lines
= 10C2
= 10!/8!2!
= 45

6.
(i)
The number of arrangements
= 4P4
= 4!
= 24

(ii)
(1) The couple choose their room (4P1).
(2) The rest 2 guests choose their rooms out of the rest 3 rooms (3P2).

The number of arrangements
= 4P1 x 3P2
= (4!/3!) x (3!/1!)
= 4 x 6
= 24

參考: 賣女孩的火柴

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