Download PERMUTATIONS and COMBINATIONS

PERMUTATIONS and COMBINATIONS
Achievement Standard 90643 (3.3) (part)
Key Words: factorial, fundamental theorem of counting, permutations combinations.
1. FACTORIALS.
Sigma 150, Ex 8.01
For a positive integer n, n ! = n(n-1)(n-2)….3.2.1
Nulake p134
n ! is called n factorial,
0! = 1
Example: 6! = 6.5.4.3.2.1
= 720 (there is a button on your calculator that finds factorials)
2. THE FUNDAMENTAL THEOREM OF COUNTING.
Example 1: If a menu offers 2 different entrees, 4 different mains and 3 different
desserts then the total number of possible meals is 243 = 24
In general if a task can be performed in n1 ways, a second task in n2 ways and a third
task in n3 ways ……….., then the total number of distinct ways of performing all tasks
together is n1 n2 n3……………
Example 2: How many 4 or 5 digit telephone numbers are possible, assuming the first
is not zero?
ans 9  10  10  10 + 9  10  10  10  10 = 99 000
3. PERMUTATIONS.
A permutation is an arrangement of objects – order is important
Sigma p154, Ex 8.02
Nulake p136
Example1: If I have 3 books (a, b, c) then there are 6
permutations of 2 of them.
ab, ba, ac, ca, bc, cb.
There are 6 possible ways that I could arrange 2 of them on a shelf.
The number of distinct permutations of n objects taking r at a time is : nPr =
n!
(n – r)!
Example 2: A club has 12 members. How many ways could a president, vice-president
and treasurer be appointed.
12
P3 =
12!
(12 – 3)!
= 1 320
The number of permutations of n objects using all of them is n!
Example 3: In how many ways can 5 people line up in a queue?
ans 5! = 120
Sigma p159, Ex 8.03
Nulake p141
4. COMBINATIONS.
A combination is a selection of objects – order is not important
Example 1: If I have 3 books (a, b, c) then there are 3 combinations of 2 of them.
ab, ac, bc.
There are 3 possible ways that I could select 2 of them.
The number of different combinations of n objects taking r at a time is :
n!
n
C = n =
r
r
(n – r)!r!
Example 2: From a club of 12 members, how many ways are there of selecting a
committee of three?
12
C3 =
12!
(12 – 3)!3!
= 220
Example 3: A group consists of 8 boys and 5 girls:
(a) How many ways can you select 2 boys and 2 girls?
(b) How many ways can you select a committee of 4 containing at least
2 boys?
ans
(a) 8C2  5C2 = 28  10
= 280
(b)
At least 2 boys = 2 boys and 2 girls + 3 boys and 1 girl + 4 boys
= 8C2  5C2 + 8C3  5C1 +8C4  5C0
= 280 + 280 + 70
= 630
5. PROBABILITY CALCULATIONS USING COMBINATIONS / PERMUTATIONS
Sigma p161, Ex 8.04, 8.05
Nulake p141
Example 1: 4 chocolates are chosen at random from a box containing 6 with hard
centres, and 8 with soft centres.
(a) Calculate the probability that 3 of the chocolates have soft centres.
(b) Calculate the probability that at least 3 of the chocolates have soft
centres.
(a) Total number of ways of selecting 4 chocolates = 14C4
= 1 001
Number of ways of selecting 3 soft centres (and 1 hard) = 8C3  6C1
= 336
P( 3 soft) = 333
1001
or 0.3357
(b) Numbers of ways of selecting at least 3 soft = 8C3  6C1 + 8C4  6C0
= 336 + 70
= 406
P( at least 3 soft) =
406
1001
or 0.4056
Example 2: A 4 digit security number is made using the digits 0, 1, ………..9.
If a number is made up at random, what is the probability that it contains
the same digit repeated 3 times in a row.
(a) Total number of security codes = 104 = 10 000.
(b) Total number of ways of getting 3 of the same in a row:
10  ( 1  1  1  9 + 9  1  1  1) = 180
(c)
P(3 in a row) =
180
10000
= 0·018