Counting for Quest 2 Do Now Class Examples 3-18. 3. Suppose you take 4 different routes to Trenton, the 3 different routes to Philadelphia. How many different.

Counting for Quest 2

Do Now

Class Examples 3-18

3. Suppose you take 4 different routes to Trenton, the 3 different routes to Philadelphia.

How many different

routes can you take

for the trip to

Philadelphia by way

of Trenton?

________ • _________

Trenton Philadelphia

___4____ • ___3_____

12

4. You have 10 pairs of pants, 6 shirts, and 3 jackets.

How many outfits

can you have

consisting of a

shirt, a pair of

pants, and a

jacket?

______•______•______

Shirts Pants Jackets

___6__•__10__•__3___

180

5. Fifteen people line up for concert tickets.

a) How many

different

arrangements are

possible?

__•__•__•__•__•__•__•__•__

•__•__•__•__•__• _=

15•14•13•12•11•10•9•8•7•6•5•4•3•2•1 =

1,307,674,368,000

b) Suppose that a

certain person must

be first and another

person must be last.

How many

arrangements are now

possible?

1 •__•__•__•__•__•__•__•__

•__•__•__•__•__• 1 =

1•13•12•11•10•9•8•7•6•5•4•3•2•1•1 =

6,227,020,800

6) Using the letters A, B, C, D, E, Fa) How many “words”can be made using all 6letters? No repetition6 • 5 • 4 • 3 • 2 • 1 = 720b) How many of thesewords begin with E ?1 • 5 • 4 • 3 • 2 • 1 = 120c) How many of thesewords do NOT beginwith E? 720 –120 = 600d) How many 4-letterwords can be made ifno repetition is allowed?6•5•4•3 = 360

e) How many 3-letterwords can be made ifrepetition is allowed?6 • 6 • 6 = 216f) How many 2 OR 3letter words can bemade if repetition isnot allowed? 6•5+6•5•4 = 30 + 120 = 150g) If no repetition isallowed, how manywords containing atleast 5 letters can bemade? (both letter 6a)720 + 720 = 1440

6) Using the letters A, B, C, D, E, Fa) How many “words”

can be made using all 6

letters? No repetition

6P6 = 6 • 5 • 4 • 3 • 2 • 1 = 720

b) How many of these

words begin with E ?

1 • 5 • 4 • 3 • 2 • 1 = 120

c) How many of these

words do NOT begin

with E? 720 –120 = 600

d) How many 4-letter

words can be made if

no repetition is allowed?

6P4 = 6•5•4•3 = 360

e) How many 3-letter

words can be made if

repetition is allowed?

6 • 6 • 6 = 216

f) How many 2 OR 3 letter

words can be made if

repetition is not allowed?

6P2 + 6P3 =

6•5 + 6•5•4 = 30 + 120 = 150

g) If no repetition is allowed,

how many words containing

at least 5 letters can be made

6P5 + 6P6 =

720 + 720 = 1440

7. How many distinguishable permutations can be made using all the letters of:

a) GREAT

__•__•__•__•__

5 • 4 • 3 • 2 • 1

5!

120

b) FOOD

4!

2!

4 • 3 • 2!

2!

12

c) TENNESSEE

9!_________

4! 2! 2!1!

9 • 8 • 7 • 6 • 5 • 4!

4! 2 • 2

15,120

4

3,780

8. Suppose you have 3 red flags, 5 green flags, 2 yellow flags, and 1 white flag. Using all the flags in a row, how many distinguishable signals can be sent?

11! =

3! 5! 2!1!

11 • 10 • 9 • 8 • 7 • 6 • 5! =

3 • 2 • 5! • 2

332,640 =

12

27,720

9. How many ways can 7 people be seated in a circle?

(7-1)! =

720

10. If you have a dozen different flowers and wish to arrange them so there is one in the center and the rest in a circle around them, how many arrangements are possible? 12 • (11-1)! =

Center Circle

12 • 3,628,800 =

43,545,600

11. Note: zero can never be the first digit of a “__-digit number”.

a) How many 4-

digit numbers

contain no nines?

__ • __ • __ • __ 8 • 9 • 9 • 9 =

5832

b) How many 4-

digit numbers contain

AT LEAST ONE nine?

__ • __ • __ • __ 9 • 10 • 10 • 10 –

8 • 9 • 9 • 9 =

9000 – 5832 =

3168

12. How many 10-letter words can you make if no letter can be repeated?

Set up using the

fundamental counting

principle.

__ • __ • __ • __ • __ •__

• __ • __ • __ • __

26•25•24•23•22•21•20•

19•18•17 =

1,927,522,397,000

Then using

permutation notation

26 P10 =

26! =

(26 – 10)!

26!

16!26•25•24•23•22•21•20•19•18•17•16!

16!

13. How many 26-letter words can be made

if no repetition of a letter is allowed?

26!

14) How ways can your homeroom (of 23 people) choose an ASC rep and a ASC alternate?

23 P2 =

23 • 22 =

506

15) Suppose we just want to select 2 people in the homeroom to serve on the ASC committee. How many 2-person groups are possible

23 C2 =

23! =

21! 2!

23 • 22 =

2

253

16) How many 5-card “hands” are possible when dealt from a

deck of 52 cards?

52 C5 =

52! =

47! 5!52 • 51 • 50 • 49 • 48 • 47! =

47! • 5 • 4 • 3 • 2 • 1

2,598,960

17. Eight points are located on the circumference of a circle.

You want to draw a triangle whosevertices are each one of these points.How many triangles are possible?_______ • _______Starting CircleVertex___7!____ • ___6!____ 5040 • 7203,628,800

18) Out of a class of 6 seniors and 5 juniors. I need to select a dance committee that must contain 2 seniors and 1 junior. How many different ways can this be done?

6 C2 • 5 C1 =

6! • 5! =

4! 2! 4! 1!

6 • 5 •4! • 5 • 4! =

4! 2 4!

75