Exercises on Combinations and Repetitions1

A - Nine cards, each of a different colour, are to be arranged in a line. How many different arrangements of the 9 cards are possible? The 9 cards include a pink card and a green card. How many different arrangements do not have the pink card next to the green card? Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line. How many different arrangements in total of 3 cards are possible? How many of the arrangements of 3 cards in part (iii) contain the pink card? How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card? Answers A - Answers and explanations Here we just need to multiply the series of descending natural numbers of 9, in other words using the factorial function. 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880 This issue can be solved by reversing the question like this: The total number of arrangements that do have the pink (P) card next to the green card (G), we assume both card are considered as one, attached together indeed! Therefore there are two possiblities: pink (P) card next to green card (G): PG = 8! = 40,320 green (G) card next to pink (P) card: GP = 8! = 40,320 >> P next to G = 2 × 8! = 80,640 Now working out the arrangements do not have the pink card next to the green card or P away from G P away from G = total number of arrangments − P next to G ==> P away from G = 9! − 2 × 8! = 362,880 − 80,640 = 282,240 Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line. How many different arrangements in total of 3 cards are possible? 9P3 = 9! (9 − 3)!  =  9! 6!  =  504 How many of the arrangements of 3 cards in part (iii) contain the pink card? With the restriction, there must be a pink (P) card, the two other cards are randomly selected and arranged from the 8 cards remaining. - 1 pink (P) out of 3 cards: ==> 3P1 = 3! 1!  =  3! = 6 ways or possibilities - 2 cards out of 8: ==> 8P2 = 8! (8 − 2)!  =  8! 6!  =  56 ways Then, there are 56 × 6 = 168 arrangements of 3 cards containing the pink card. How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card? Again, we can reverse the question: Firstly, we work out "The arrangements of 3 cards that do have the pink card next to the green card and green next to pink": PG or GP - 2 × 2! = 4 ways - 1 card randomly select out of the 7 remaining: 7 ways >> 4 × 7 = 28 ways P aways G = 504 (see (iii)) Therefore: The arrangements of 3 cards that do not have the pink card next to the green card is: 504 − 28 = 476