# Sigma solutions Permutations and Combinations. Ex. 8.03 Page 159.

## Presentation on theme: "Sigma solutions Permutations and Combinations. Ex. 8.03 Page 159."— Presentation transcript:

Sigma solutions Permutations and Combinations

Ex. 8.03 Page 159

Sigma: Page 159 Ex 8.03 6. Museum: 149 Maori artifacts. Selects 3 each month. How many selections poss? (a)In factorial form? So (b)As an ordinary number? 149 C 3 = 540 274 possible selections.

Sigma: Ex 8.03 cont. 7. Couple has 13 friends; invites a different 4 each week. How long to the nearest year till all possible groups selected? Selecting 4 from 13 each week. Number poss. selections = 13 C 4 = 715 So 715 weeks till all possible groups of 4 invited. 52 weeks in a year. Nbr years = 715 ÷ 52 = 13.75 = 14 years (to nearest yr).

Sigma: Ex 8.03 cont. 8. There are 12 shirts on sale. A man wants to buy 5. (a)How many poss. choices? Nbr poss. selections = 12 C 5 = 792 choices. (b)If 2 are too expensive? Now selecting 5 from only 10 (2 ruled out). Nbr poss. selections = 10 C 5 = 252 choices. (c)If he damages 1 so must buy it? He MUST buy the 1 he damaged (decision made), so now selecting 4 from the other 11. Nbr poss. selections = 11 C 4 = 330 choices.

9. How many selections of at least 2 objects can be made from 5? AT LEAST 2 means selections of 2, 3, 4 or 5 objects. Nbr selections = 5 C 2 + 5 C 3 + 5 C 4 + 5 C 5 = 26 possible selections. 10. 6 points on a circle. Any 2 can be joined. (a)How many different chords can be formed? There can be a chord between ANY 2 of the 6 points. Think of as pairs of points (i.e. selections of 2 items from 6). Draw a circle and try it. What the Q is really asking: How many possible selections of 2 can be made from the 6 points? Answer: 6 C 2 = 15 possible chords. (b)How many possible segments? Each chord in a circle forms 2 segments – one on each side. So nbr of segments = 2 × nbr of chords = 2 × 15 = 30 segments. Sigma: Ex 8.03 cont.

11. There are 7 candidates. Can vote for up to 3 candidates for 3 vacancies. In how many ways can a vote be cast? Can vote for UP TO 3 candidates, so can also vote for just 1 candidate or for 2 candidates. Nbr poss. selections of 1, 2 or 3 from 7 = 7 C 1 + 7 C 2 + 7 C 3 = 63 ways of casting a valid vote. Sigma: Ex 8.03 cont.

12. Elections are to be held for the School Common Room Committee. Five students are to be elected. The candidates are six female students and seven male students. (a)How many distinct committees can be elected? There are 13 candidates in total. In how many ways can 5 be selected? 13 C 5 = 1287 distinct committees. (b)How many distinct committees with two female and three male students can be elected? There are 6 C 2 ways of selecting 2 female students from 6. There are 7 C 3 ways of selecting 3 male students from 7. Nbr poss. selections with 2F & 3M students = 6 C 2 × 7 C 3 = 525 distinct committees. Sigma: Ex 8.03 cont.

13. A class of 12 students and a teacher are to go on a class trip. A car will take four of the party, and the rest will go by bus. In how many different ways can they all go on the trip if: (a)The teacher must go by car? The teacher is in the car (fixed) so 3 spots in car remaining: So were selecting 3 of the 12 students to go in the car with the teacher. Number of possible combinations = 12 C 3 (same as 12 C 9 for bus) = 220 (b)The teacher must go by bus? The teacher is in the bus (fixed) so 8 spots in the bus remaining: So were selecting 8 of the 12 students to go in the bus with the teacher. Number of possible combinations = 12 C 8 (same as 12 C 4 for bus) = 495 Sigma: Ex 8.03 cont.

13. A class of 12 students and a teacher are to go on a class trip. A car will take four of the party, and the rest will go by bus. In how many different ways can they all go on the trip if: … (b)The teacher must go by bus? The teacher is in the bus (fixed) so 8 spots in the bus remaining: So were selecting 8 of the 12 students to go in the bus with the teacher. Number of possible combinations = 12 C 8 (same as 12 C 4 for bus) = 495 (c)The teacher can go by either car or bus? Nbr. poss. combos = nbr combos if teacher is in car + if teacher if on bus = 12 C 3 (from (a)) + 12 C 8 (from (b)) = 220 + 495 = 715 possible combinations. Sigma: Ex 8.03 cont.

14. A buffet restaurant has a selection of six salads, four cold meat platters and seven cooked vegetables. How many ways are there for a person to choose three salads, two types of cold meat and four vegetables? There are 6 C 3 ways of selecting 3 salads from 6. There are 4 C 2 ways of selecting 2 types of cold meat from 4. There are 7 C 4 ways of selecting 4 types of cooked vegetables from 7. Nbr poss. selections= 6 C 3 × 4 C 2 × 7 C 4 =20 × 6 × 35 =4200 possible choices. Sigma: Ex 8.03 cont.

15. * (Scholarship type problem) In the game of badminton a singles match has two players – one against another; and a doubles match has four players – two players against two others. If eight friends play badminton, how many different doubles and singles matches can be played? Number of possible singles matches= number of possible selections of 2 from 8. = 8 C 2 = 28 possible singles matches Number of possible doubles matches = number of possible selections of 4 from 8 × number of possible doubles games played by each group of 4. = 8 C 4 × 4 C 2 / 2 * = 210 possible doubles matches *Q:Why divide by 2? Sigma: Ex 8.03 cont.

15. * (Scholarship type problem) In the game of badminton a singles match has two players – one against another; and a doubles match has four players – two players against two others. If eight friends play badminton, how many different doubles and singles matches can be played? Number of possible singles matches= number of possible selections of 2 from 8. = 8 C 2 = 28 possible singles matches Number of possible doubles matches = number of possible selections of 4 from 8 × number of possible doubles games played by each group of 4. = 8 C 4 × 4 C 2 / 2 * = 210 possible doubles matches *Q:Why divide by 2?

15. * (Scholarship type problem) In the game of badminton a singles match has two players – one against another; and a doubles match has four players – two players against two others. If eight friends play badminton, how many different doubles and singles matches can be played? Number of possible singles matches= number of possible selections of 2 from 8. = 8 C 2 = 28 possible singles matches Number of possible doubles matches = number of possible selections of 4 from 8 × number of possible doubles games played by each group of 4. = 8 C 4 × 4 C 2 / 2 * = 210 possible doubles matches *Q:Why divide by 2? A: Because whenever you select one pair from each 4, you are also selecting another of the pairs – the other 2. Hence by going 4 C 2 we actually get twice the number of possible doubles games.

16. (Australian Mathematics Comp. type problem) 12 people decide to play volleyball. In how many ways can they be split up into two equal teams? Number of possible teams= (number of possible selections of 6 from 12) / 2. * = 12 C 6 / 2 = 924 / 2 = 462 possible ways of splitting into 2 teams of 6. * Q: Why divide by 2? A: Because the question asks how many ways there are off splitting the 12 people into 2 teams, not how many possible teams. Every time you split it, you get 2 teams. Sigma: Ex 8.03 cont.

17. (Australian Mathematics Comp. type problem) In how many ways can 10 different items be split up into two unequal sets? Number of possible unequal sets (i.e. every possible number except 5) = 10 C 1 + 10 C 2 + 10 C 3 + 10 C 4 * = 10 + 45 + 120 + 210 = 385 possible ways of splitting them into two unequal sets. *sets of 6 or more are already covered - e.g. you automatically select a set of 6 when you select 4 from 10, and a set of 7 when you select 3 from 10 etc. Sigma: Ex 8.03 cont.

18. The Department of Foreign Affairs has trained 30 linguists, 16 of whom are to be assigned to Germany and 14 to France. (a)(NCEA Merit level) If 8 of the linguists refuse to go to France, in how many ways can the 30 linguists be assigned? The 8 who wont go to France MUST go to Germany (fixed). So there are 22 people left. Must select 8 for the remaining German positions. By doing this you automatically select the other 14 for France. Number of possible ways of selecting 8 from 22 to fill the remaining German positions = 22 C 8 (same answer if you go 22 C 14 for France) = 319 770 ways. (b)(NCEA Excellence level) If, in addition to the 8 who refuse to go to France, 10 refuse to go to Germany, how many ways are there of assigning the 30 linguists? Sigma: Ex 8.03 cont.

18. (NCEA Excellence level) The Department of Foreign Affairs has trained 30 linguists, 16 of whom are to be assigned to Germany and 14 to France. … (b)If, in addition to the 8 who refuse to go to France, 10 refuse to go to Germany, how many ways are there of assigning the 30 linguists? The 8 who wont go to France MUST go to Germany (fixed). The 10 who wont go to Germany MUST go to France (fixed). So there are 12 people left (30 – 8 – 10 = 12). Must select 8 for the remaining German positions. By doing this you automatically select the other 4 for France. Number of possible ways of selecting 8 from 12 to fill the remaining German positions = 12 C 8 (same answer if you go 12 C 4 for France) = 495 ways. Sigma: Ex 8.03 cont.

Similar presentations