Presentation is loading. Please wait.

Presentation is loading. Please wait.

Sigma solutions Permutations and Combinations. Ex. 8.02 Page 154.

Similar presentations


Presentation on theme: "Sigma solutions Permutations and Combinations. Ex. 8.02 Page 154."— Presentation transcript:

1 Sigma solutions Permutations and Combinations

2 Ex Page 154

3 Sigma: Page 154 Ex Horse trainer: 9 horses, 4 jockeys. 9 P 4 = 3024 ways. 3. EQUATIONS: How many words with 4 letters, each used just once. 9 P 4 = 3024 ways. 4. Hotel room: 4 people assigned to 16 different rooms. 16 P 4 = ways. 5. RHOMBUS: How many different 5-letter words. Last letter always u. u is fixed at the end ( _ _ _ _ u ), so think of it as: In how many ways can you arrange RHOMBS into 4- letter words? 6 P 4 = 360 ways

4 Ex 8.02 cont. 6. Gambler A bets on winner & 2 nd ; Gambler B bets on winner, 2 nd & 3 rd. 18 horses in race Bets in $1 units. Gambler A outlay = $1 × nbr possible arrangements of 1 st, 2 nd. = $1 × 18 P 2 = $306 Gambler B outlay = $1 × nbr possible arrangements of 1 st, 2 nd, 3 rd. = $1 × 18 P 3 = $4896

5 Ex 8.02 cont. 7. In how many ways can 4 bollards & a child be arranged if: (a) The child is on one end? The child could be on either end, then, for each of those, the 4 bollards could be arranged in 4! different ways: Nbr arrangements of bollards × Nbr possible positions for child = 4 P 4 × 2 =4! × 2 =48 ways. OR, using the multiplication principle: 2×4×3×2×1×1 = 48 ways. (b) The child is NOT on one end? Must be a bollard on both ends then. Child could be in any of the three positions in between. Nbr arrangements of bollards × Nbr possible positions for child = 4 P 4 × 3 =4! × 3 =72 ways.

6 Ex 8.02 cont. 8. How many words with 3 letters can be formed from NORMAL without repeats, given that? (a) The letter n must be used? n is guaranteed so dont need to select it, but it could be in 3 possible positions. Nbr ways = Number of possible arrangements of 2 letters from O R M A L × nbr possible positions for n = 5 P 2 × 3 = 60 ways (b) The letter n must not be used? Nbr ways = Number of possible arrangements of 3 letters from O R M A L = 5 P 3 = 60 ways

7 Ex 8.02 cont girls & 4 boys standing in line. How many ways of arranging them if: (a) They can stand in any position? Number of ways = Number of possible arrangements of all 9 = 9 P 9 or 9! = ways (b) Shortest girl stands at one end? Number of ways = Number of possible positions for shortest girl × number of poss arrangements of the other 8 people = 2 × 8! (the 2 is because the girl could be at either end) = ways (c) Each pair of girls has a boy in between (alternate)? Must be GBGBGBGBG so, using the multiplication principle: Number of ways = 5 × 4 × 4 × 3 × 3 × 2 × 2 × 1 × 1 = 5! × 4! = 2880 ways.

8 Ex 8.02 cont. 9. cont. (d) The 5 girls stand together? The options are:

9 Ex 8.02 cont. 9. cont. (d) The 5 girls stand together? The options are: GGGGGBBBB

10 Ex 8.02 cont. 9. cont. (d) The 5 girls stand together? The options are: GGGGGBBBB =>5! × 4! ways. BGGGGGBBB => 5! × 4! ways. BBGGGGGBB =>5! × 4! ways. BBBGGGGGB =>5! × 4! ways. BBBBGGGGG =>5! × 4! ways. So total number of poss arrngmts = 5 ×5! × 4! = ways

11 Ex 8.02 cont people line up. How many possible arrangements if shortest cant be next to tallest? Answer = Number of ways of arranging 8 people – nbr arrgmts where shortest is beside tallest. So first need to find number of orders where shortest is beside tallest: To find this, treat them as an inseparable pair. So now its like there are 7 people (6 people + an inseparable pair (shortest & tallest). There are 7! ways of arranging 7 people. However, this must be doubled. Why? Because there are 2 possible orders for our inseparable pair: shortest then tallest, or tallest then shortest. So nbr arrngmts where shortest is beside tallest = 2 × 7! Answer = Number of ways of arranging 8 people – nbr arrgmts where shortest is beside tallest. = 8! – 2 × 7! = ways.

12 Ex 8.02 cont. 11. In how many ways can a rowing 8 be made up from 4 North Islanders and 4 South Islanders? (a)If there are no restrictions on seating? Nbr ways = 8 P 8 = 8! = ways (b) If all the South Islanders sit at the front? Nbr ways = Nbr ways of arnging 4 SI ers at front × Nbr ways of arnging 4 NI ers at back. = 4! × 4! = 576 ways (c)If the North and South Islanders must alternate seats? Must be either: NSNSNSNS or SNSNSNSN Total nbr ways= nbr ways of getting NSNSNSNS + nbr ways SNSNSNSN = 4! × 4! + 4! × 4! = 1152 ways.

13 Ex 8.02 cont. 12. How many different 8-letter words can be formed from the letters of the word binomial? B I N O M I A L 8 letters so all possible arrangements would be 8! (i.e. 8 P 8 ). However there are 2 Is. All the letters must be used for each arrangement (8-letter words), so both Is will be present in all 8! possible arrangement. Think of it as B I 1 N O M I 2 A L 8! is the number of arrangements if the order of the 2 Is (i.e. which comes first and which comes second) is relevant. In reality this is irrelevant – it makes no difference which I occurs first in the word. Theyre both the same! (e.g. B I 1 N O M I 2 A L is the same as B I 2 N O M I 1 A L). We must therefore halve the number of arrangements, so the final answer is: Nbr of ways of arranging BINOMIAL into an 8-letter word without consecutive Is

14 Ex 8.02 cont. 12. How many different 8-letter words can be formed from the letters of the word binomial? B I N O M I A L 8 letters so all possible arrangements would be 8! (i.e. 8 P 8 ). However there are 2 Is. All the letters must be used for each arrangement (8-letter words), so both Is will be present in all 8! possible arrangement. Think of it as B I 1 N O M I 2 A L 8! is the number of arrangements if the order of the 2 Is (i.e. which comes first and which comes second) is relevant. In reality this is irrelevant – it makes no difference which I occurs first in the word. Theyre both the same! (e.g. B I 1 N O M I 2 A L is the same as B I 2 N O M I 1 A L). We must therefore halve the number of arrangements, so the final answer is: Nbr of ways of arranging BINOMIAL into an 8-letter word = 8! ÷ 2 = ways.

15 Ex 8.02 cont. *13. (Excellence level) How many ways can a path be laid in a straight line using 7 brick paving stones, 4 concrete slabs and 5 slate tiles (all are the same size)? Total nbr of ways = = = ways.

16 Ex 8.02 cont. 14. Gareth is arranging his magazines in order on the shelf: 3 surfing mags (S), 7 computer mags (C), 5 rugby league mags (R). In how many ways can the mags be arranged if: (a)There are no restrictions? Were just arranging 15 different mags, regardless of type. This is different from Q13 because (unlike the paving stones) every item is different (i.e. no 2 magazines are the same).

17 Ex 8.02 cont. 14. Gareth is arranging his magazines in order on the shelf: 3 surfing mags (S), 7 computer mags (C), 5 rugby league mags (R). In how many ways can the mags be arranged if: (a)There are no restrictions? Were just arranging 15 different mags, regardless of type. This is different from Q13 because (unlike the paving stones) every item is different (i.e. no 2 magazines are the same). Thus we can simply treat the 15 magazines as 15 different objects being arranged. We can forget about type (i.e. S, C or R). Answer: 15! (i.e. 15 P 15 ) = × ways (to 3dp)

18 Ex 8.02 cont. 14. cont. (b) The magazines of each type stay together as a group? The options are:

19 Ex 8.02 cont. 14. cont. (b) The magazines of each type stay together as a group? The options are: 3S, 7C, 5R =>3! × 7! × 5! ways. 3S, 5R, 7C =>3! × 7! × 5! ways. 7C, 3S, 5R =>3! × 7! × 5! ways. 7C, 5R, 3S =>3! × 7! × 5! ways. 5R, 3S, 7C =>3! × 7! × 5! ways. 5R, 7C, 3S=>3! × 7! × 5! ways. So there are 6 (or 3!) different possible orders of the 3 groups. So total nbr of arngmts possible=6 × 3! × 7! × 5! = ways

20 Ex 8.02 cont. 15. Christmas street parade: 3 Pipe Bands, 5 Floats & 2 Clowns. (a)How many possible arrangements for the parade? Like with 14(a), were just arranging all of the objects, regardless of type. In this case it is 10 different items in a parade. We dont need to worry about type (i.e. bands, floats, or clowns). Answer: 10! (i.e. 10 P 10 ) = ways. (b) The pipe bands comes first & the clowns are placed last?

21 Ex 8.02 cont. 15. Christmas street parade: 3 Pipe Bands, 5 Floats & 2 Clowns. (a)How many possible arrangements for the parade? Like with 14(a), were just arranging all of the objects, regardless of type. In this case it is 10 different items in a parade. We dont need to worry about type (i.e. bands, floats, or clowns). Answer: 10! (i.e. 10 P 10 ) = ways. (b) The pipe bands comes first & the clowns are placed last? Must be: 3P, 5F, 2C

22 Ex 8.02 cont. 15. Christmas street parade: 3 Pipe Bands, 5 Floats & 2 Clowns. (a)How many possible arrangements for the parade? Like with 14(a), were just arranging all of the objects, regardless of type. In this case it is 10 different items in a parade. We dont need to worry about type (i.e. bands, floats, or clowns). Answer: 10! (i.e. 10 P 10 ) = ways. (b) The pipe bands comes first & the clowns are placed last? Must be: 3P, 5F, 2C = 3! × 5! × 2! ways. = 1440 ways


Download ppt "Sigma solutions Permutations and Combinations. Ex. 8.02 Page 154."

Similar presentations


Ads by Google