Probability and Counting: Passwords, Combinations, and Geometry, Study notes of Cryptography and System Security

Various counting techniques and probability concepts, including password possibilities, three-letter combinations, seating arrangements, and geometric probabilities. It covers topics such as constructive counting, complementary counting, and combinations with and without repetition.

Typology: Study notes

Pre 2010

Uploaded on 09/17/2009

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PROBABILITY
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27
pf28
pf29
pf2a
pf2b
pf2c
pf2d

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PROBABILITY

Counting

Multiple Stages With replacement

  • HOW MANY PASSWORD ARE POSSIBLE STARTING WITH TWO LETTERS FOLLOWED BY FOUR NUMERALS?
  • (26)(26)(10)(10)(10)(10) = 6,760,

WITHOUT REPLACEMENT

  • How many three letter words from { a,b,c,d,e} without using a letter more than once?

With restrictions

  • How many three letter words from { a,b,c,d,e} without using a letter more than once?
  • How many three letter words from { a,b,c,d,e} without using a letter more than once and a vowel in the middle?
  • Ans. (4)(2)(3) = 24
  • How many three digit numbers have one zero?
  • Case 1 zero in the middle (9)(1)(9) = 81
  • Case 2 zero on the right (9)(9)(1) = 81
  • Total 81 +81 =

Complementary

(finding opposite first)

  • How many ways to seat four boys and three girls in a row so that at least two boys sit together?

Multiple methods

  • In how many ways may 3 officers (Pres. ,V.P. , Sec) be picked from a 12 member club, If Al and Bob won’t serve together?
  • In how many ways may 3 officers (Pres. ,V.P. , Sec) be picked from a 12 member club, If Al and Bob won’t serve together?
  • Ans. Ways to pick three of 12: (12)(11)(10)

  • Ways to pick three with Al and Bob: (3)(2)(10) =60. Thus 1320-60 = 1260 ways to pick without both.
  • How many ways tho arrange the letters of Mississippi?
  • Ans. There are 11! ways to arrange the letters. There are 4! ways to arrange the “s” and “I”, and 2! For “p”. Thus 11!/[(4!)(4!)(2!) =34650 diferent arrangements.

Symmetry

  • How many ways can 6 people be seated at a round table. It does not matter who is seated north. What matters is who is on a persons left and right.

Combinations

(order does not matter)

  • In How many ways can a committee of four be picked from a club of ten.
  • In How many ways can a committee of four be picked from a club of ten.
  • Ans. There are (10)(9)(8)(7) way to pick, but there are 4! Ways to arrange each pick so there are duplications. Thus there are (10)(9)(8)(7)/4! Different committees. Combinations are represented by 10 C4.