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.

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Presentation transcript:

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? __________________ __________ _= = 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 = = 6,227,020,800

6) Using the letters A, B, C, D, E, F a)How many “words” can be made using all 6 letters? No repetition = 720 b)How many of these words begin with E ? = 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? 6543 = 360 e) How many 3-letter words can be made if repetition is allowed? = 216 f) How many 2 OR 3 letter words can be made if repetition is not allowed? = = 150 g) If no repetition is allowed, how many words containing at least 5 letters can be made? (both letter 6a) = 1440

6) Using the letters A, B, C, D, E, F a)How many “words” can be made using all 6 letters? No repetition 6 P 6 = = 720 b)How many of these words begin with E ? = 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? 6 P 4 = 6543 = 360 e) How many 3-letter words can be made if repetition is allowed? = 216 f) How many 2 OR 3 letter words can be made if repetition is not allowed? 6 P P 3 = = = 150 g) If no repetition is allowed, how many words containing at least 5 letters can be made 6 P P 6 = = 1440

7. How many distinguishable permutations can be made using all the letters of: a)GREAT __________ ! 120 b)FOOD 4! 2! 4 3 2! 2! 12 c)TENNESSEE 9!_________ 4! 2! 2!1! ! 4! , ,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! ! = 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? __ __ = 5832 b) How many 4- digit numbers contain AT LEAST ONE nine? __ __ – = 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. __ __ __ __ __ __ __ = 1,927,522,397,000 Then using permutation notation 26 P 10 = 26! = (26 – 10)! 26! 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 P 2 = = 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 C 2 = 23! = 21! 2! = 2 253

16) How many 5-card “hands” are possible when dealt from a deck of 52 cards? 52 C 5 = 52! = 47! 5! ! = 47! ,598,960

17. Eight points are located on the circumference of a circle. You want to draw a triangle whose vertices are each one of these points. How many triangles are possible? _______ Starting Circle Vertex ___7!____ ___6!____ ,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 C 2 5 C 1 = 6! 5! = 4! 2! 4! 1! 6 5 4! 5 4! = 4! 2 4! 75