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Sigma solutions

Sigma solutions. Permutations and Combinations. Ex. 8.02. Page 154. Sigma: Page 154 Ex 8.02. 2. 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 .

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Sigma solutions

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  1. Sigma solutions Permutations and Combinations

  2. Ex. 8.02 Page 154

  3. Sigma: Page 154Ex 8.02 2. Horse trainer: 9 horses, 4 jockeys. 9P4 = 3024 ways. 3. EQUATIONS: How many words with 4 letters, each used just once. 9P4 = 3024 ways. 4. Hotel room: 4 people assigned to 16 different rooms. 16P4 = 43 680 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? 6P4 = 360 ways

  4. Ex 8.02 cont. 6. Gambler A bets on winner & 2nd; Gambler B bets on winner, 2nd & 3rd. 18 horses in race Bets in $1 units. Gambler A outlay = $1 × nbr possible arrangements of 1st, 2nd. = $1 × 18P2 = $306 Gambler B outlay = $1 × nbr possible arrangements of 1st, 2nd, 3rd. = $1 × 18P3 = $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 = 4P4 × 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 = 4P4 × 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 don’t 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 = 5P2× 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 = 5P3 = 60 ways

  7. Ex 8.02 cont. 9. 5 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 = 9P9 or 9! = 362 880 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) = 80 640 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! = 14 400 ways

  11. Ex 8.02 cont. 10. 8 people line up. How many possible arrangements if shortest can’t 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 it’s 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! = 30 240 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? • If there are no restrictions on seating? Nbr ways = 8P8 = 8! = 40 320 ways (b) If all the South Islanders sit at the front? Nbr ways = Nbr ways of arnging 4 SIers at front × Nbr ways of arnging 4 NIers at back. = 4! × 4! = 576 ways • If the North and South Islanders must alternate seats? Must be either: NSNSNSNSor 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. 8P8). • However there are 2 I’s. All the letters must be used for each arrangement (8-letter words), so both I’s will be present in all 8! possible arrangement. • Think of it as B I1 N O M I2 A L • 8! is the number of arrangements if the order of the 2 I’s (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. They’re both the same! (e.g. B I1 N O M I2 A L is the same as B I2 N O M I1 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 I’s

  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. 8P8). • However there are 2 I’s. All the letters must be used for each arrangement (8-letter words), so both I’s will be present in all 8! possible arrangement. • Think of it as B I1 N O M I2 A L • 8! is the number of arrangements if the order of the 2 I’s (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. They’re both the same! (e.g. B I1 N O M I2 A L is the same as B I2 N O M I1 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 = 20 160 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 = = =1 441 440 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: • There are no restrictions? We’re 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: • There are no restrictions? We’re 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. 15P15) = 1.308 × 1012 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! = 21 772 800 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), we’re just arranging all of the objects, regardless of type. In this case it is 10 different items in a parade. We don’t need to worry about type (i.e. bands, floats, or clowns). Answer: 10! (i.e. 10P10) = 3 628 800 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), we’re just arranging all of the objects, regardless of type. In this case it is 10 different items in a parade. We don’t need to worry about type (i.e. bands, floats, or clowns). Answer: 10! (i.e. 10P10) = 3 628 800 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), we’re just arranging all of the objects, regardless of type. In this case it is 10 different items in a parade. We don’t need to worry about type (i.e. bands, floats, or clowns). Answer: 10! (i.e. 10P10) = 3 628 800 ways. (b) The pipe bands comes first & the clowns are placed last? Must be: 3P, 5F, 2C = 3! × 5! × 2! ways. = 1440 ways

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