Homework 3

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 1^st gentleman can sit with 10 possible ladies

The 2^nd gentleman can sit with 9 possible ladies

…

The 9^th gentleman can sit with 2 possible ladies

The 10^th gentleman can sit with 1 possible ladies

Total possible pairs = 10! = 3628800

b.

The 1^st lady can sit with 19 possible ladies

The 2^nd lady can sit with 18 possible ladies

…

The 9^th lady can sit with 11 possible ladies

The 10^th 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 6^th 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