**Question 1**:

The diagram below shows five cards of different letters.

**(a)**Calculate the number of arrangements, in a row, of all the cards.

**(b)**Calculate the number of these arrangements in which the letters

*E*and

*A*are side by side.

*Solution:***(a)**Number of arrangements = 5! =

**120**

**(b)**

*Step 1*If the letters ‘

*E*’ and ‘

*A*’ have to be placed side by side, they will be considered as one item.

Together with the letters ‘

*R*’, ‘

*C*’ and ‘

*T*’, there are altogether 4 items.

__EA__

__R__

__C__

__T__

**4!**

Step 2Step 2

*E*’ and ‘

*A*’ can also be arranged among themselves in their group.

Number of arrangements =

**2!**

Hence, number of arrangements of all the letters of the word ‘

*REACT*’ in which the letters

*E*and

*A*have to be side by side

= 4! × 2!

= 24 × 2

=

**48**

**:**

Question 2

Question 2

A group of 4 men and 3 ladies are to be seated in a row for a photographing session. If the men and ladies want to be seated alternately (man-lady-man-lady…), calculate the number of different arrangements.

*Solution:*The arrangements of 4 men and 3 ladies to be seated alternately are as follow:

__M__

__L__

__M__

__L__

__M__

__L__

__M__

The number of ways to arrange the seat for 4 men =

**4!**

The number of ways to arrange the seat for 3 ladies =

**3!**

Total number of different arrangements for men and ladies = 4! × 3! =

**144**

**Question 3**:

Ahmad has 6 durians, 5 watermelons and 2 papayas. If he wants to arrange these fruits in a row and the fruits of the same kind have to be grouped together, calculate the number of different arrangements. The sizes of the fruits are different.

*Solution:*The number of ways to arrange 3 groups of fruits that are same kind =

**3!**

__DDDDDD__

__WWWWW__

__PP__

The number of ways to arrange 6 durians =

**6!**

The number of ways to arrange 5 watermelons =

**5!**

The number of ways to arrange 2 papayas =

**2!**

Therefore, the number of ways to arrange the fruits with same kind of fruits was grouped together

= 3! × 6! × 5! × 2!

=

**1036800**

**Question 4**:

Calculate the number of arrangements, without repetitions, of the letters from the word `SOMETHING’ with the condition that they must begin with a vowel.

*Solution:*$\begin{array}{l}\text{Arrangementoflettersbeginwithvowel}\\ \text{=}O,\text{}M\text{and}I\\ ={\text{}}^{3}{P}_{1}\\ \\ \text{Arrangementfortherestoftheletters}\\ \text{=}7!\\ \\ \therefore \text{Numberofarrangements}\\ {\text{=}}^{3}{P}_{1}\times 7!\\ =15120\end{array}$