Sigma solutions Permutations and Combinations. Ex. 8.04 Page 161

Sigma solutions

Permutations and Combinations

Ex. 8.04

Page 161

Sigma: Page 161Ex 8.04

1. Bank allocates 4-digit PIN numbers. Digits chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9}. No digit repeated.

(a)How many different PIN numbers are possible?

Without replacement and order matters - arranging (not just selecting). permutations.

? ? ? ?





? ? ? ?





9 ? ? ?

9 possibilities for the first digit





9 ? ? ?





9 8 ? ?

8 possibilities for the second digit – one used up





9 8 ? ?





9 8 7 ?

7 possibilities for the third digit – 2 used up





9 8 7 ?





9 8 7 6

6 possibilities for the third digit – 3 used up





)!(

!n

rn

nPr

= 3024 different PINS are possible.

!5

!94

9 P Same as saying: 9 × 8 × 7 × 6

9 8 7 6


…(b)How many of the PIN numbers start with an odd digit?


? ? ? ?



Number of possibilities for first digit = 5


? ? ? ?



Number of possibilities for first digit = 5 (i.e. 1, 3, 5, 7 or 9)


5 ? ? ?





5 ? ? ?




Then there are 3 remaining positions to fill.


5 ? ? ?




Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement.


5 ? ? ?






5 ? ? ?






5 8 ? ?






5 8 ? ?






5 8 7 ?






5 8 7 ?






5 8 7 6




Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations.


5 8 7 6




Then there are 3 remaining positions to fill. 8 unused digits left to choose from without replacement. Order matters so still permutations. i.e. 8P3

Nbr poss. PINS starting with an odd digit = 5 × 8P3


5 8 7 6






or 5 × 8 × 7 × 6


5 8 7 6






or 5 × 8 × 7 × 6 = 1680 PINS.


5 8 7 6


…(c) Calculate the probability that a client is given a PIN

number starting with an odd digit

Nbr poss. PINS starting with an odd digit (from b) = 5 × 8P3

= 1680

Sigma: Page 161 - Ex 8.04

Total number poss. PIN numbers (from a) = 9P4

= 3024

P(1st digit is odd) = numbers PIN poss.number Total

digit oddan with starting PINS poss.Nbr

3024

1680 =

9

5 = answer

Sigma: Page 161 - Ex 8.042. Company ‘numbers’ its invoices to clients using 1 letter of

the alphabet followed by 3 digits from 0 to 9, repeats allowed.

(a)How many different ‘numbers’ could be used?




Repeats allowed (i.e. with replacement) so the number of possibilities does not reduce after a digit is used.

Letter

? ? ? ?

3 digits from 0 to 9





Letter

? ? ? ?






Letter

26 ? ? ?






Letter

26 ? ? ?






Letter

26 10 ? ?






Letter

26 10 10 ?






Letter

26 10 10 10


2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed.



= 26 000 different ‘numbers’ possible.


Letter

26 10 10 10


So number of possible ‘numbers’ = 26 × 10 × 10 × 10

or 26 × 103



…(b)How many would end in the numbers 8 or 9?

Repeats still allowed (i.e. with replacement) so the number of possibilities for the other positions stays the same.

Letter

26 10 10 ?

2 digits from 0 to 9 Last digit is either 8 or 9



…(b)How many would end in the numbers 8 or 9?

Repeats still allowed (i.e. with replacement) so the number of possibilities for the other positions stays the same.

Letter

26 10 10 2

2 digits from 0 to 9 Last digit is either 8 or 9

2 possibilities

So number of possible ‘numbers’ = 26 × 10 × 10 × 2

or 26 × 102 × 2

= 5 200 different ‘numbers’ possible ending in 8 or 9.

2. Company ‘numbers’ its invoices to clients using 1 letter of the alphabet followed by 3 digits from 0 to 9, repeats allowed.

(c) Calculate the probability of choosing an invoice that does not contain the number 1 (NCEA Merit level).

Nbr poss. invoice numbers not containing a 1 = 26 × 93

= 18 954


Total number poss. invoice ‘numbers’ (from a) = 26 × 103

= 26 000

P(does not contain a 1) = numbers invoice poss.number Total

1 a containingnot numbers invoice poss.Nbr

26000

18954 =

= 0.729 answer

5. A cricket team of 11 is to be chosen from a squad of 14, 2 of whom are brothers. Find the probability that:

(a) Both brothers are chosen.

There is only one way of selecting both brothers: 2C2 - i.e. 1.


Total number poss. combinations for the 11 = 14C11

= 364

P(selecting both brothers in the 11) = 11 for the nscombinatio poss. ofnumber Total

12 from brothers-non 9 selecting of waysNbr.

1114

912

C

C =

= or 0.6044 (4sf) answer

This is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations.

Number of ways of selecting the other 9 from the other 12 = 12C9 - i.e. 220.

91

55


(b) Neither brother is chosen.

There is only one way of selecting neither brother: 2C0 - i.e. 1.



= 364

P(selecting neither bro. in the 11) = 11 for the nscombinatio poss. ofnumber Total

12 from brothers-non 11 selecting of waysNbr.

1114

1112

C

C =


Still is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations.

Nbr ways of selecting all 11 from the 12 non-brothers = 12C11 - i.e. 12.

91

3


(c) Only one brother is chosen. (NCEA Merit level)

Nbr ways of selecting 1 brother from 2: 2C1 - i.e. 2 (could pick either bro).



= 364

P(selecting one bro. in the 11) = 11 for the nscombinatio poss. ofnumber Total

bros-non 10 selecting of Ways 1bro selecting of Ways

1114

1012

12

C

CC =


Still is a selection-only problem (note the word ‘chosen’). Order is not important. So we’re dealing with Combinations.

Nbr ways of selecting 10 from the 12 non-brothers = 12C10 - i.e. 66.

91

33

6. (a) Complete the probability distribution table for the nbr of red balls obtained when 3 balls are selected at random from an urn containing 6 red and 4 yellow balls. (NCEA Merit level)

P(X=0)


P(selecting 0 red balls) = balls 3 of selections poss. ofnumber Total

yellow 3 selecting of Ways red 0 selecting of Ways

310

34

06

C

CC =

=

Nbr ways of selecting 3 of the 4 yellow balls = 4C3 - i.e. 4.

30

1

Nbr ways of selecting 0 of the 6 red balls = 6C0 - i.e. 1.

Let X: number of red balls selected.

x 0 1 2 3

P(X=x)30

1


P(X=1)


P(selecting 1 red ball) = balls 3 of selections poss. ofnumber Total


310

24

16

C

CC =

=


10

3



x 0 1 2 3

P(X=x)30

1

30

9

30

9

or

x 0 1 2 3

P(X=x)


P(X=2)


P(selecting 2 red ball) = balls 3 of selections poss. ofnumber Total


310

14

26

C

CC =

=


2

1



30

1

30

15

30

9

or

30

15

x 0 1 2 3

P(X=x)


P(X=3)


P(selecting 3 red balls) = balls 3 of selections poss. ofnumber Total


310

04

36

C

CC =

=


6

1



30

1

30

5

30

9

or

30

15

30

5

x 0 1 2 3

P(X=x)

6. (b) If X is the number of red balls obtained, find an expression for P(X=x). The expression should use combinations.

(NCEA Excellence level)

P(X= x)


P(selecting x red balls) = balls 3 of selections poss. ofnumber Total

yellow x)-(3 selecting of Ways red x selecting of Ways

310

)3(46

C

CC xx = answer

Nbr ways of selecting (3- x) of the 4 yellow balls = 4C(3-x)

Nbr ways of selecting x of the 6 red balls = 6Cx


30

130

9

30

15

30

5

7. Three people each have a pair of socks, which are all washed one day then returned at random. What is the probability that a particular person gets their own pair of socks back?


P(selecting a given pair) = 6 from socks 2any selecting of waysofnumber Total

pair specific a selecting of Ways

26

22

C

C =

=

Nbr ways of selecting a specific pair (selecting 2 socks from 2) = 2C2 - i.e. 1.

15

1

Total number of ways of selecting any 2 socks from the 6 = 6C2 - i.e. 15.

Selection only. Order doesn’t matter (left sock same as right sock). combinations.


8. In the card game of Poker, a hand of 5 cards is dealt from a pack of 52. If all 5 cards are from the same suit, the hand is called a ‘flush’. What is the probability of getting a ‘flush’ in the game of Poker?

Without replacement. Selecting only so order doesn’t matter. combinations.

Total nbr of possible selections of ANY 5 cards = 52C5

= 2 598 960

Number of possible ‘flushes’ – i.e. selections of 5 of same suit:





= 2 598 960


= 4 × 13C5





= 2 598 960


= 4 × 13C5 (i.e. 4 suits – each has 13 cards and we’re choosing 5 of them)

= 5148 possible selections of 5 of same suit.




= 2 598 960


= 4 × 13C5 (i.e. 4 suits – each has 13 cards)

= 5148 possible selections of 5 of same suit.

P(getting a flush) = cards 5 ANY of selections possible ofNumber

suit same theof 5 of selections possible ofNumber

552

5134

C

C =

= or 0.001981 (4sf) answer 16660

33



OR can solve the problem without using combinations, just by using the multiplication principle.

Total possible number of arrangements of any 5 cards:

52 51 50 49 48× × × ×





52 51 50 49 48× × × ×




Number of possible ‘flushes’ – i.e. ways of getting 5 of same suit:


? ? ? ? ?

52 51 50 49 48× × × ×


? ? ? ? ?

52 possibilities for the first card (could be any card)





52 51 50 49 48× × × ×


52 ? ? ? ?

52 possibilities for the first card (could be any card)





52 51 50 49 48× × × ×






52 ? ? ? ?

52 51 50 49 48× × × ×

12 possibilities for the second card (the other 12 from the same suit)






52 12 ? ? ?

52 51 50 49 48× × × ×

12 possibilities for the second card (the other 12 from the same suit)






52 12 ? ? ?

52 51 50 49 48× × × ×

11 possibilities for the 3rd card (2 used up, so 11 of that suit are remaining)






52 12 11 ? ?

52 51 50 49 48× × × ×

11 possibilities for the 3rd card (2 used up, so 11 of that suit are remaining)






52 12 11 ? ?

52 51 50 49 48× × × ×

10 possibilities for the 4th card (3 used up, so 10 of that suit are remaining)






52 12 11 10 ?

52 51 50 49 48× × × ×







52 12 11 10 ?

52 51 50 49 48× × × ×







52 12 11 10 9

52 51 50 49 48× × × ×







52 12 11 10 9

52 51 50 49 48× × × ×







52 51 50 49 48× × × ×

52 12 11 10 9× × × ×




52 51 50 49 48× × × ×

52 12 11 10 9× × × ×P(getting a flush) =

cards 5 of tsarrangemen poss. ofnumber Total

flushes'' possible ofNumber

4849505152

910111252

=

= or 0.001981 (4sf) answer 16660

33

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Sigma solutions Permutations and Combinations. Ex. 8.04 Page 161