Permutation and Combination Questions HELP (1 Viewer)

[csi]

1. In how many ways can a committee of 3 women and 4 men be chosen from 8 women and 7 men if two particular women refuse to serve on the committee together?
ANSWER: 1750

2. Eight points are equally spaced on the circumference of a circle. Chords may be drawn by joining any two points. How many chords can be drawn which are not diameter?
ANSWER: 24

3. Three identical pens and four identical pencils are arranged in a row. If only five writing implements are used, how many arrangements are possible?
ANSWER: 25

4. In how many ways can 6 different books be distributed between 2 students, provided that both students have at least 1 book.
ANSWER: 62

5. Bottles are coded by different arrangements of coloured squares in a row. The colours used are green, red and yellow. The maximum number of green squares is 3, the maximum of red ones two and at most one only is yellow.
(a) How many different codes are possible if 6 squares are used?
ANSWER: 60
(b) If only 5 squares are used, how many different codes are possible?
ANSWER: 60

Thanks!

 

[cossine]

Q1.
let women be A and B

case 1: A is on the committee

6C2

case 2: B is on the committee

6C2

case 3: neither A, B is on the committee

6C3

7C4( 6C2 + 6C2 + 6C3) = 1750

Q2.

8*6/2! = 24 (8 options at first, since can't consider the point opposite there then 6 options. Divide by 2! as the order is not important)

Q3.

let the 3 pens be called pen A and the other pen B

case 1: 0 pen A

not possible

case 2: 1 pen A

5C1 (nCk works by providing the permutations as it is dealing with the repetition of 4 pens B and 1 pen A. The nCk used in binomial distribution also outputs the number permuations in the same manner.)

case 3: 2 pen A

5C2

case 4: 3 pen A

5C3

no other cases are possible

5C1 + 5C2 + 5C3 = 25


Q4

Consider all cases for 1 person

person 1 has:

1 book => 6 options

2 books => 6*5 /2!

3 books => 6*5*4 /3!

4 books => 6*5*4*3/4!

5 books=> 6*5*4*3*2/5!

Add all cases

6 + 15 + 20 + 15 + 6 = 62


Q5 a)

6!/ ((3!)*2!*1!) = 60


b) Through intuition it possible to tell the answer is 60. This is because there is only one choice for non-existent square 6.

But you can consider each case as a double-check.

 

[csi]

Q1.
let women be A and B

case 1: A is on the committee

6C2

case 2: B is on the committee

6C2

case 3: neither A, B is on the committee

6C3

7C4( 6C2 + 6C2 + 6C3) = 1750

Q2.

8*6/2! = 24 (8 options at first, since can't consider the point opposite there then 6 options. Divide by 2! as the order is not important)

Q3.

let the 3 pens be called pen A and the other pen B

case 1: 0 pen A

not possible

case 2: 1 pen A

5C1 (nCk works by providing the permutations as it is dealing with the repetition of 4 pens B and 1 pen A. The nCk in binomial distribution also outputs the number permuations in the same manner.)

case 3: 2 pen A

5C2

case 4: 3 pen A

5C3

no other cases are possible

5C1 + 5C2 + 5C3 = 25


Q4

Consider all cases for 1 person

person 1 has:

1 book => 6 options

2 books => 6*5 /2!

3 books => 6*5*4 /3!

4 books => 6*5*4*3/4!

5 books=> 6*5*4*3*2/5!

Add all cases

6 + 15 + 20 + 15 + 6 = 62


Q5 a)

6!/ ((3!)*2!*1!) = 60


b) Through intuition it possible to tell the answer is 60. This is because there is only one choice for non-existent square 6.

But you can consider each case as a double-check.

tysm