Question

1. each of the following counting problems are worth 2 marks each - show all of your work!a) how many ways are there to make a sandwich with egg salad or chicken salad, lettuce or tomato, butter or mayonnaise, on a whole wheat bun, a whole grain bagel, or a sesame seed bun?b) how many four - digit passwords can be made from the digits 0 to 5, if digits can be repeated?c) how many ways are there to draw a heart or queen from a standard deck of 52 cards?d) how many arrangements are there using all letters of the word cucumber?

Explanation:

Step1: <Apply the fundamental counting principle>

For part a), the number of choices for salad: \(2\) (egg or chicken). The number of choices for vegetable: \(2\) (lettuce or tomato). The number of choices for spread: \(2\) (butter or mayonnaise). The number of choices for bun: \(3\) (whole - wheat, whole - grain bagel, sesame - seed). By the fundamental counting principle \(N = 2\times2\times2\times3\).

Step2: <Calculate the product>

\(2\times2\times2\times3=24\).

For part b), the first digit (thousands place) of the four - digit password: Since it's a four - digit password, the first digit (thousands place) can't be \(0\). So, there are \(5\) choices (\(1 - 5\)). The second, third, and fourth digits (hundreds, tens, and units place) can be any of the \(6\) digits (\(0 - 5\)). By the fundamental counting principle \(N=5\times6\times6\times6\).

Step3: <Calculate the product>

\(5\times6\times6\times6 = 5\times216=1080\).

For part c), use the formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Let \(A\) be the set of heart cards and \(B\) be the set of Queen cards. In a standard deck of \(52\) cards, \(n(A) = 13\) (13 heart cards), \(n(B)=4\) (4 Queen cards: Queen of hearts, diamonds, clubs, spades), and \(n(A\cap B) = 1\) (Queen of hearts). Then \(n(A\cup B)=13 + 4-1\).

Step4: <Calculate the sum and difference>

\(13 + 4-1=16\).

For part d), the word \(CUCUMBER\) has \(8\) letters. The letter \(C\) appears \(2\) times and the letter \(U\) appears \(2\) times. The formula for permutations of a multi - set is \(\frac{n!}{n_1!n_2!\cdots n_k!}\), where \(n\) is the total number of objects, and \(n_i\) are the number of times each repeated object appears. Here \(n = 8\), \(n_1=2\) (for \(C\)) and \(n_2 = 2\) (for \(U\)). So \(N=\frac{8!}{2!2!}\).

Step5: <Calculate the factorial and quotient>

\(8! = 8\times7\times6\times5\times4\times3\times2\times1=40320\), \(2! = 2\times1 = 2\). Then \(\frac{8!}{2!2!}=\frac{40320}{2\times2}=10080\).

Answer:

a) \(24\)
b) \(1080\)
c) \(16\)
d) \(10080\)