perms and combs help!! (1 Viewer)

koolkid59

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PLZ do all these questions with all working out
i just can't get them!


1) If you toss two coins and roll three dice, how many outcomes are possible?

2) Jack has six different football cards and Meg has another eight different football cards. In how many ways can one of Jack's cards be exchanged for one of Meg's cards?

3) In Sydney, phone numbers at present consist of eight digits, starting with the digit 9.
(d) How many are there that do not contain a zero, and in which the consecutive digits alternate between odd and even?

4)
(a) If repetitions are not allowed, how many four-digit numbers can be formed from the digits 1, 2, ... 8, 9?
(c) How many of these are even?
(e) How many are greater than 7000?

5) Find how many arrangements of the letters of the word UNIFORM are possible:
(a) if the vowels must occupy the first, middle and last positions
(d) if the M is somewhere to the right of the U.

6. In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?

7. In Morse code, letters are formed by a sequence of dashes and dots. How many different letters is it possible to represent if a maximum of ten symbols are used?

8. Five-letter words are formed without repetition from the letters of PHYSICAL.
(d) How many contain the letter Y?

9.
(a) Integers are formed from the digits 2, 3, 4 and 5, with repetitions not allowed.
(ii) How many of them are even?
(b) Repeat the two parts to this question if repetitions are allowed.

10. Five backpackers arrive in a city where there are five youth hostels.
(c) Suppose that two of the backpackers are brother and sister and wish to stay in the same youth hostel. How many different accommodation arrangements are there if the other three can go to any of the other youth hostels?
 

someth1ng

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I don't guarantee that all my answers are correct as I do make mistakes but these are my answers:

1) If you toss two coins and roll three dice, how many outcomes are possible?
n=2x2x6x6x6=864 outcomes

2) Jack has six different football cards and Meg has another eight different football cards. In how many ways can one of Jack's cards be exchanged for one of Meg's cards?
n=6x8=48 ways

3) In Sydney, phone numbers at present consist of eight digits, starting with the digit 9.
(d) How many are there that do not contain a zero, and in which the consecutive digits alternate between odd and even?
1,2,3,4,5,6,7,8,9
n in form 9EOEOEOE=4x5x4x5x4x5x4=32000 numbers

4)
(a) If repetitions are not allowed, how many four-digit numbers can be formed from the digits 1, 2, ... 8, 9?
n=9P4=3024 numbers
(c) How many of these are even?
Even numbers will end in 2,4,6 or 8 which is 4/9 of the numbers. 3024x(4/9)=1344 even numbers
(e) How many are greater than 7000?
>7000 will start with 7,8 or 9. 3024x(3/9)=1008 numbers

5) Find how many arrangements of the letters of the word UNIFORM are possible:
(a) if the vowels must occupy the first, middle and last positions
VCCVCCV - Arranging the Vowels=3! and Arranging the Consonants=4!
Total Arrangements=3!4!=144 Arrangements
(d) if the M is somewhere to the right of the U.
apollo1's Method=7!/2!=2520 ways

6. In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?
n=(8C3)(5C2)=560 ways

7. In Morse code, letters are formed by a sequence of dashes and dots. How many different letters is it possible to represent if a maximum of ten symbols are used?
One Symbol=2^1
Two Symbols=2^2
Three Symbols=2^3
Continue until Ten Symbols
.`. Total Letters=2+4+8+16...2^10=2046 letters

8. Five-letter words are formed without repetition from the letters of PHYSICAL.
(d) How many contain the letter Y?
Containing Y=(7C4)(5!)=4200 words

9.
(a) Integers are formed from the digits 2, 3, 4 and 5, with repetitions not allowed.
(ii) How many of them are even?
Even numbers will end in 2 or 4.
Total numbers=4!=24 numbers
Even numbers=(1/2)(24)=12 numbers
(b) Repeat the two parts to this question if repetitions are allowed.
Total Numbers=4x4x4x4=256 numbers
Even numbers=(1/2)(256)=128 numbers

10. Five backpackers arrive in a city where there are five youth hostels.
(c) Suppose that two of the backpackers are brother and sister and wish to stay in the same youth hostel. How many different accommodation arrangements are there if the other three can go to any of the other youth hostels?
The pair will pick a hostel so there are 5 ways.
The other 3 can pick out of 4 hostels so there is 4x4x4 ways

Total Arrangements=5x4x4x4=320 ways
 
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apollo1

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5 (d) the M is somewhere to the right of the U.

u can think of this is cases where M is the first letter, M is second and so on.

or another way is to place the M and place the U. the arrangements of the rest are 7!. but since each arrangement has a distinct alternate arrangement you divide by 2!

therefore 7!/2! = 2520.

i used to use the cases method before largarithmic (A bos user) taught me about the second method which is much easier and quicker in an exam.
 

someth1ng

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another way is to place the M and place the U. the arrangements of the rest are 7!. but since each arrangement has a distinct alternate arrangement you divide by 2!

therefore 7!/2! = 2520.

i used to use the cases method before largarithmic (A bos user) taught me about the second method which is much easier and quicker in an exam.
I don't understand this method...explain?

EDIT: Actually, I understand this method. So if it was asking for 3 letters in XZY order, would it be divided by 3! and so on?
 
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Kyrix

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I don't guarantee that all my answers are correct as I do make mistakes but these are my answers:

1) If you toss two coins and roll three dice, how many outcomes are possible?
n=2x2x6x6x6=864 outcomes

2) Jack has six different football cards and Meg has another eight different football cards. In how many ways can one of Jack's cards be exchanged for one of Meg's cards?
n=6x8=48 ways

3) In Sydney, phone numbers at present consist of eight digits, starting with the digit 9.
(d) How many are there that do not contain a zero, and in which the consecutive digits alternate between odd and even?
1,2,3,4,5,6,7,8,9
n in form 9EOEOEOE=4x5x4x5x4x5x4=32000 numbers

4)
(a) If repetitions are not allowed, how many four-digit numbers can be formed from the digits 1, 2, ... 8, 9?
n=9P4=3024 numbers
(c) How many of these are even?
Even numbers will end in 2,4,6 or 8 which is 4/9 of the numbers. 3024x(4/9)=1344 even numbers
(e) How many are greater than 7000?
>7000 will start with 7,8 or 9. 3024x(3/9)=1008 numbers

5) Find how many arrangements of the letters of the word UNIFORM are possible:
(a) if the vowels must occupy the first, middle and last positions
VCCVCCV - Arranging the Vowels=3! and Arranging the Consonants=4!
Total Arrangements=3!4!=144 Arrangements
(d) if the M is somewhere to the right of the U.
apollo1's Method=7!/2!=2520 ways

6. In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?
n=(8C3)(5C2)=560 ways

7. In Morse code, letters are formed by a sequence of dashes and dots. How many different letters is it possible to represent if a maximum of ten symbols are used?
One Symbol=2^1
Two Symbols=2^2
Three Symbols=2^3
Continue until Ten Symbols
.`. Total Letters=2+4+8+16...2^10=2046 letters

8. Five-letter words are formed without repetition from the letters of PHYSICAL.
(d) How many contain the letter Y?
Containing Y=(7C4)(5!)=4200 words

9.
(a) Integers are formed from the digits 2, 3, 4 and 5, with repetitions not allowed.
(ii) How many of them are even?
Even numbers will end in 2 or 4.
Total numbers=4!=24 numbers
Even numbers=(1/2)(24)=12 numbers
(b) Repeat the two parts to this question if repetitions are allowed.
Total Numbers=4x4x4x4=256 numbers
Even numbers=(1/2)(256)=128 numbers

10. Five backpackers arrive in a city where there are five youth hostels.
(c) Suppose that two of the backpackers are brother and sister and wish to stay in the same youth hostel. How many different accommodation arrangements are there if the other three can go to any of the other youth hostels?
The pair will pick a hostel so there are 5 ways.
The other 3 can pick out of 4 hostels so there is 4x4x4 ways

Total Arrangements=5x4x4x4=320 ways
I think questions 6 and 9 were done wrongly.

For 6, it says in how many ways it can be arranged, which means it is unordered so combinations cant be used.
So number of ways of placing particular women on bow side = 4x3x2
Number of ways of placing women on stroke = 4x3
Rest of women = 3!
Total no. = 3! x 4 x 3 x 4 x 3 x 2 = 1728

For 9, I think you forgot to consider the fact that i can be 1 digit, two digit, etc..

for 9 (a)

1 digit = 2 ways
2 digit = 3 x 2 ways = 6
3 digit = 2x2x3 ways = 12
4 digit = 1 x2 x2 x 3 = 12
Total = 2+ 6 + 12 + 12 = 32 ways

(ii)
1 didgit = 2
2 dig = 4x 2 = 8
3 dig = 4x4 x2 = 32
4 dig = 4x4x4 x 2 = 128
Total = 170 ways
 

Kyrix

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I don't understand this method...explain?

EDIT: Actually, I understand this method. So if it was asking for 3 letters in XZY order, would it be divided by 3! and so on?

I think it should work for 3 objects, but not too sure and i dont know why

because for two objects say ..X...Y..
In the total number of arrangements X can either be infront of Y or Y can be infront of X
So number of ways of half = no of arrangements / 2

For three Objects i usually do be safe

Set XYZ in place
X Y Z
Another object can be placed in 4 different places
*X* Y* Z*
Adding another object can be placed in 5 different places now
*X*(1st object)*Y*Z*
another object 6 places... etc.
Total number of ways = 4x5x ...
 

someth1ng

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I think questions 6 and 9 were done wrongly.

For 6, it says in how many ways it can be arranged, which means it is unordered so combinations cant be used.
So number of ways of placing particular women on bow side = 4x3x2
Number of ways of placing women on stroke = 4x3
Rest of women = 3!
Total no. = 3! x 4 x 3 x 4 x 3 x 2 = 1728
I think this would depend if you consider the all spots on the stroke side to be different/same and same for bow side. The question doesn't specifically state this. I was under the impression that it was meant by "arrangements" as in how many ways you can have three on the stroke, two on the bow side and the remaining people since I suppose that's still arranging them into groups (parts of the boat).

For 9, I think you forgot to consider the fact that i can be 1 digit, two digit, etc..
In this one, I believe when they said 2,3,4 AND 5 meaning that all 4 digits were needed to be used. By the "with repetitions not allowed", I thought that was to force it to use the 4 digits.

Note: I highly doubt that ambiguities like this will arise in the HSC since the questions are reviewed to ensure accuracy. I believe this is from a textbook which is not reviewed to such an extent like the HSC exams are so these issues do occur.
 
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