The document discusses permutations and combinations. It defines a permutation as an arrangement of objects where order matters, and a combination as an arrangement where order does not matter. It provides examples of permutations of sets with 3 and 8 elements. The number of permutations of a set of n elements is n factorial. Permutations are arrangements of all elements of a set. The document also provides examples of calculating the number of permutations in different scenarios like seating arrangements and rearrangements of letters in a word.

4. If the order doesn't matter, it is a Combination.
If the order does matter it is a Permutation.
there are six permutations of the set:
{1,2,3},
(1,2,3), (1,3,2), (2,1,3),
(2,3,1), (3,1,2), (3,2,1).

5. The number of permutations of n distinct
objects is n factorial usually written as
“n! “, which means the product of all
positive integers less than or equal to
”n”.

7. A permutation of a set S is defined as
a bijection from S to itself. The collection
of such permutations form a symmetric
group. Permutations may act on
structured objects by rearranging their
components, or by certain replacements
(substitutions) of symbols.

8. the r-permutations, or partial
permutations, are the ordered
arrangements of r distinct elements
selected from a set. When r is equal to
the size of the set, these are the
permutations of the set.

9. Example 1:
n! = n(n-1)(n-2)…1
9! = 9*8*7*6*5*4*3*2*1
answer = 362,880
By definition:
1! = 1 and 0! = 1
Formula:
n!

10.  How many ways can 8 students be seated in a row of 6
chairs?
 Solve by (a) Multiplication rule & (b) Permutation of
n things taken r at time.
FORMULA:
nPr = n!
(n-r)!

11. Notation:
Example 2:
Given:
n = 8, r = 6
Solution:
= 8! = 8! = 40320 = 20,160
(8 – 6)! 2! 2
or
Manual = 8*7*6*5*4*3*2! ÷ 2!
= 20,160
Formula:
nPr = n!
(n – r)!

12.  How many ways can 8 people sit around a round table?
FORMULA:
(n-1)!

13. Example 3:
Given:
n = 8
Solution:
(n – 1)! = (8 – 1) = 7!
= 7*6*5*4*3*2*1!
answer = 5,040
Formula:
circular = (n – 1)!

14.  In how many ways can the letters of the word
“ MISSISSIPPI” be arranged?
FORMULA:
n!
n1! n2! n3!...nk!

15. Example 4:
Given:
from the word “MISSISSIPPI”
Solution: Formula:
n = 11 n1 / n1! n2! … nk!
n1 = 1 (M)
n2 = 4 (I)
n3 = 4 (S)
n4 = 2 (P)
= 11! = 11*10*9*8*7*6*5*4*3*2*1
1! 4! 4! 2! (1)(4.3.2.1)(4.3.2.1)(2.1)
= (11.10)(9.7)(5)
= (110)(63)(5)
= (6930)(5)
Answer = 34650