Question 1:

lower + upper case = 26 x 2 = 52

all characters = 52 + 10 + 1 = 63

Assuming that the minimum number of characters can be one, then:

Number of 1 character variables = 52 x 63 0

Number of 2 character variables = 52 x 63 1

Number of 3 character variables = 52 x 63 2

Number of 4 character variables = 52 x 63 3

Number of 5 character variables = 52 x 63 4

Number of 12 character variables = 52 x 63 11

Total = 3.27867 x 10 21

Question 2:

(a). There are 5! (= 120) anagrams

valid words are: aster, rates, stare,

(b). Allowing repeated letters should give more possibilities.

(c).

i.

Number of 1 character words = 5

Number of 2 character words = 5 x 4

Number of 3 character words = 5 x 4 x 3

Number of 4 character words = 5 x 4 x 3 x 2

Number of 5 character words = 5 x 4 x 3 x 2 x 1

Total = 325

Increase = (325 – 120) / 120 x 100% = 171 %

ii.

Total = 5 5 = 3125

increase = (3125 – 120) / 120 x 100% = 2504 %

Question 3:

(a). Total possible keys = 2 56 = 7.20576 x 10 16

(b). Total time = (7.20576 x 10 16 / (10 9 /sec x 3.154 x 10 7 sec/year) = 2.28464 years

(c). Total possible keys = 2 168 = 3.74144 x 10 50

Total time = (3.74144 x 10 50 / (10 9 /sec x 3.154 x 10 7 sec/year) = 1.18625 x 10 34 years

Question 4:

(a).

number of recipes with 0 mixins = 15

number of recipes with 1 mixins = 15 x (27! / 26!)

number of recipes with 2 mixins = 15 x (27! / 25!)

number of recipes with 26 mixins = 15 x (27! / 1!)

number of recipes with 27 mixins = 15 x 27!

Total = 4.43985 x 10 29

(b). Total with at least 2 = Total – num with 0 mixins – num with 1 mixin

≈ Total = 4.43985 x 10 29

Question 5:

x = number of bits needed

2 x >= 100,000

log 2 (2 x ) >= log 2 (100,000)

x >= log 2 (100,000) = 16.6

x = 17 bits

Question 6:

The 1st gentleman can sit with 10 possible ladies

The 2nd gentleman can sit with 9 possible ladies

The 9th gentleman can sit with 2 possible ladies

The 10th gentleman can sit with 1 possible ladies

Total possible pairs = 10! = 3628800

b.

The 1st lady can sit with 19 possible ladies

The 2nd lady can sit with 18 possible ladies

The 9th lady can sit with 11 possible ladies

The 10th lady can sit with 10 possible ladies

Total possible pairs = 19! / 9! = 3.352212864 x 10 11

Question 7:

(a).

Breakfast choices = 3

Lunch choices = 4

Dinner choices = 3

Total menus = 3 x 4 x 3 = 48

(b).

Breakfast splurge options = 1

Lunch splurge options = 2

Dinner splurge options = 1

Breakfast only options = 1 x (4-2) x (3-1)

Lunch only options = (3-1) x 2 x (3-1)

Dinner only options = (3-1) x (4-2) x 1

Total menus = 1x2x3 + 2x2x2 + 2x2x1 = 18

Question 8:

Assumptions:

Total number of core classes still needed = 17

Total number of core classes that prerequisites have been met = 13

Total number of elective classes still needed = 3

Total number of eligible elective classes = 9

(add one more option in 6th course for no course)

schedules with 1 electives = 9 x 13 x 12 x 11 x 10 x (9 + 1) = 1,544,400

schedules with 2 electives = 9 x 8 x 13 x 12 x 11 x (10 + 1) = 1,359,072

schedules with 3 electives = 9 x 8 x 7 x 13 x 12 x (11 + 1) = 943,488

Total = 3,846,960