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

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