Sigma solutions
Permutations and Combinations
Ex. 8.02
Page 154
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
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
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.
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
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.
Ex 8.02 cont.
9. cont.
(d) The 5 girls stand together?
The options are:
Ex 8.02 cont.
9. cont.
(d) The 5 girls stand together?
The options are: GGGGGBBBB
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
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.
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 = 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
(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.
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 BINOMIALinto an 8-letter word without consecutive I’s
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 BINOMIALinto an 8-letter word = 8! ÷ 2
= 20 160 ways.
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)?
eir types within thslabs thearranging of waysofNumber
slabs 16 all arranging of waysofNbr Total nbr of ways =
=!5!4!7
!16
= 1 441 440 ways.
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?
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).
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?
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)
Ex 8.02 cont.
14. cont.(b) The magazines of each type stay together as a group?
The options are:
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
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?
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
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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