PERMUTATION:
- Suppose an event can be chosen in p different ways. Another independent event can be chosen in q different ways. Then the two events can be chosen successively in p.q ways.
Example Problem:
On
a trip a man took 3 suits, 2 ties, and 2 hats. How many different
choices of these items of clothing are possible:
Answer:
There
are 3.2.2 = 12 different ways.
- The number of permutations of n objects, taken r at a time, is defined by:
P(n,r)
= n! / (n-r)!
Example Problem:
How
many different three-letter letter patterns can be formed using the
letters a,b,c,d, and e without repetition?
Answer:
Using P(n,r)
= n! / (n-r)! formula,
P (5,3) = 5! / (5-3)! = (5.4.3.2.1) / (5-3)! = 60.
Meaning
there are 60 ways to arrange the letters a,b,c,d and e three at a
time.
- The number of permutations of n objects, taken n at a time, is defined by:
P
(n,n) = n!
Example Problem:
How many different sample of size 4 can we form the wooden block with letters A, B ,C, D?
Answer: Using P (n,n) = n! formula, P(4,4)= 4! / (4-4)! = 24
PERMUTATION WITH REPETITIONS AND CIRCULAR PERMUTATIONS
n! / p!q!
Example Problem:
How many Five-Letter patterns can be formed from the letters of the word teeth?
Answer: 5! / 2!2! = 30
PERMUTATION WITH REPETITIONS AND CIRCULAR PERMUTATIONS
- The number of Permutations of n objects of which p are alike and q are alike is found by:
n! / p!q!
Example Problem:
How many Five-Letter patterns can be formed from the letters of the word teeth?
Answer: 5! / 2!2! = 30
- Suppose n objects are arranged in a circle.Then There are n! / n or (n-1)! permutations of the n objects around the circle:
Example Problem:
A vending machine has six different items on a revolving tray. How many ways can the items be arranged on a tray?
Answer: (6-1)! = 120
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