Question 1: (3 Marks)
(a) Find the number of ways of drawing a red queen or a black king from a standard deck of playing cards. (1 Mark)
Answer: Let C denote the set of red queens and D the set of black kings. Clearly, | C |= 2 =| D | . Since C and D are disjoint sets, by the addition principle, | C ∪ D |=| C |+| D |= 2 + 2 = 4. Thus there are four different waysof drawing a red queen or a black king
(b) Write the addition principle. (1 Mark)
Answer: Let A and B be two mutually exclusive tasks. Suppose task A can be done in m ways and task B in n ways. Then task A or task B can take place in m + n ways.
(c) Write the Inclusion-Exclusive Principle. (1 Mark)
Answer: Suppose a task A can be done in m ways, task B in n ways, and both can be accomplished in k different ways. Then task A or B can be done in m + n − k ways.
Question 2:
(3 Marks)
(a) Write the multiplication principle. (1 Mark)
Answer: Suppose a task T is made up of two subtasks, subtask T 1 followed by subtask T 2 . If subtask T 1 can be done in m 1 ways and subtask T 2 in m 2 different ways for each way subtask T 1 can be done, then task T can be done in m 1 m 2 ways
(b) Find the number of two-letter words that begin with vowel – a, e, i, o, or u. (1 Mark)
Answer:
5 * 6 = 130
(c) A set S with n elements has how many subsets? (1 Mark)
Answer: it has
2n
subsets
Question 3:
(3 Marks)
(a) One type of automobile license plate number in Massachusetts consists of one letter and five digits. Calculate the number of such license plate numbers possible. (1 Mark)
Answer:
Choosing a letter can be done in 26 ways and the position has only 6 possible slots and choosing five digits 10*10*10*10*10 = 100,000
26 * 6 * 100,000 = 15,600,00
(b) An eight-bit word is called byte. Find the number of bytes with their second bit 0 or the third bit 1. (1 Mark)
Answer:
Number of bytes with second bit 0= 2 1 2 2 2 2 2 2 = 2^7
Number of bytes with second bit 1= 2 2 1 2 2 2 2 2 = 2^7
Number of bytes with second bit 0 and third bit 1 = 2 1 1 2 2 2 2 2 = 2^6
2^7 + 2^7 – 2^6 = 192
(c) Find the number of permutations, that is, 3-permutations of the elements of the set {a, b, c}. (1 Mark)
Answer:
3 * 2 * 1 = 6
Therefore P(3,3) = 6