Question

a) Freddie has 6 toys cars and 3 toy buses, all different.

i)He chooses 4 toys to take on holiday with him. In how many different ways can Freddie choose 4 toys?

ii)Freddie arranges these 9 toys in a line. Find the number of possible arrangements

•if the buses are all next to each other.

•if there is a car at each end of the line and no buses are next to each other.

b) There are 2 purple cubes, 3 blue cubes, 2 red cubes, 4 orange cubes, 2 green cubes and 2 yellow cubes. Calculate the number of possible arrangements when all the cubes are arranged in a circle.

Answer 1

(a)

(i)Total number of toys he have = 9

So, number of ways to selecting 4 toys = $^9C_4=126$

(ii) If he arranges all toys in a line

and buses are all next to each other

$[B_1\ B_2\ B_3] ,C_1,C_2,C_3,C_4,C_5,C_6$

So, number of ways to arrange = $7!\times 3!$ =30,240

Car at the end of the line but no buses are next to each other

$C_1\_\_C_2\_\_C_3\_\_C_4\_\_C_5\_\_C_6$

Buses can be filled in the gap between two toy cars

So, number of ways to arrange = $6!\times \ ^5P_3=43200$

(b) Given,

2 Purple Cubes

3 Blue cubes

2 Red cubes

4 orange Cubes

2 Green Cubes

2 Yellow Cubes

Total number of cubes = 15

If all the cubes are arranged in circle

The number of possible ways to arrange = $\dfrac{(15-1)!}{2!\times3!\times2!\times4!\times2!\times2!}=37837800$