1. • Combination: The number of ways to collect
things. Order does NOT matter.
• Permutation: An ordering of n objects is a
permutation of the objects. ORDER MATTERS!

2. Permutations vs. Combinations

3. Permutation or Combination?
• Nine books placed in a row on a shelf
Permutation
• Three books selected from a collection of 20 books.
Combination
• An arrangement of the letters in the word “BOOK”
Permutation
• Selecting the Lead actor and the understudy in a school
play.
Permutation
• Selecting a committee of 3 people from a group of 10
office workers.
Combination

4. THERE ARE 6 PERMUTATIONS OF THE
LETTERS A, B, &C
ABC ACB BAC BCA CAB CBA
THERE IS ONLY 1 COMBINATION OF
THE LETTERS A, B, C
ABC

5. IN GENERAL, THE # OF
PERMUTATIONS OF N OBJECTS IS:
n!

6. 12 SKIERS…
How many different ways can 12 skiers in the
Olympic finals finish the competition? (if there
are no ties)
12! =
12*11*10*9*8*7*6*5*4*3*2*1 =
479,001,600 different ways

7. BACK TO THE FINALS IN THE OLYMPIC
SKIING COMPETITION
How many different ways can 3 of the
skiers finish 1st, 2nd, & 3rd (gold, silver,
bronze)
Any of the 12 skiers can finish 1st, the
any of the remaining 11 can finish 2nd,
and any of the remaining 10 can finish
3rd.
So the number of ways the skiers can
win the medals is
12*11*10 = 1320

8. PERMUTATION OF N OBJECTS TAKEN R
AT A TIME
nPr =
!
!
rn
n

9. BACK TO THE LAST PROBLEM WITH
THE SKIERS
It can be set up as the number of permutations of 12
objects taken 3 at a time.
12P3 = 12! = 12! =
(12-3)! 9!
12*11*10*9*8*7*6*5*4*3*2*1 =
9*8*7*6*5*4*3*2*1
12*11*10 = 1320

10. 10 COLLEGES, YOU WANT TO VISIT
ALL OR SOME
How many ways can you visit
6 of them:
Permutation of 10 objects taken 6
at a time:
10P6 = 10!/(10-6)! = 10!/4! =
3,628,800/24 = 151,200

11. HOW MANY WAYS CAN YOU VISIT
ALL 10 OF THEM:
10P10 =
10!/(10-10)! =
10!/0!=
10! = ( 0! By definition = 1)
3,628,800

12. ONE LAST EXAMPLE - WITH REPETITION
 In how many ways can you arrange the letters of
the word Hubbub?
 Notice how the B happens 3 times, and the U
happens 2 times.
 Normally we would write 6! , but we must divide out
the b’s and u’s.
 The formula becomes:
!3!2
!6

13. COMBINATION OF N OBJECTS TAKEN R
AT A TIME
nCr =
)!(!
!
rnr
n
The formula is similar to
Permutations except we have to
divide out the duplicates.

14. FIND THE NUMBER OF COMBINATIONS OF THE
LETTERS IN THE WORD “SOLVE” TAKEN:
a. 5 at a time.
b. 2 at a time.
5 at a time: SOLVE
2 at a time: SO, SL, SV, SE
OL, OV, OE,
LV, LE
VE

15. IN HOW MANY WAYS CAN A COMMITTEE OF 6
BE CHOSEN FROM 5 TEACHERS AND 4
STUDENTS IF:
a. All are equally eligible?
b. The committee must include 3 teachers and 3
students?

16. PRACTICE
 How many combinations of the letters in the word
“ABSOLUTE” are there, taking the letters three at a
time?
 How many different five player teams can be formed
from eight people?
 In how many ways can the letters of the word
SPECIAL be arranged using:
a. All 7 letters
b. Only 4 letters at a time.
 In how many ways can you arrange the letters in the
word ILLINOIS