Permutation Short Questions (Question 1 - 4)


Question 1:
The diagram below shows five cards of different letters.
R      E      A      C      T
(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

Number of arrangements = 4!

Step 2
The letters ‘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:
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:
Arrangement of letters begin with vowel  O, E and I =   3 P 1 Arrangement for the rest of the letters 7! Number of arrangements  3 P 1 ×7! =15120