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Permutation and combination

by look416
Tags: combination, permutation
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look416
#1
Sep15-09, 11:44 AM
P: 87
1. The problem statement, all variables and given/known data
a computer terminal displaying text can generate 16 different colours numbered 1 to 16. any one of colours 1 to 8 may be used as the "background colour" on the screen, and any one of colours 1 to 16 may be used as the " text colours"; however, selecting the same colour for background and text renders the text invisible and so this combination is not used. find the number of different usable combinations of background colour and text colour.


2. Relevant equations
nCr nPr


3. The attempt at a solution
i have using
16 P 8 x 16 P 16
where 16P8 refers as the possible combinations of background colours
and 16P16 is the possible combinations of text colours
by multiplying both probability, and find the answer...
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Billy Bob
#2
Sep15-09, 01:12 PM
P: 393
Fundamental Principle of Counting: If Step 1 can be done m ways, and after it is done, Step 2 can be done n ways, then both of these steps can be done in this order in m times n ways.

Step 1 is pick one background colour. (How many background colours are there?)

Step 2 is pick one text colour.
look416
#3
Sep15-09, 02:26 PM
P: 87
Quote Quote by Billy Bob View Post
Fundamental Principle of Counting: If Step 1 can be done m ways, and after it is done, Step 2 can be done n ways, then both of these steps can be done in this order in m times n ways.

Step 1 is pick one background colour. (How many background colours are there?)

Step 2 is pick one text colour.
As what you have said,
8 background colours there , therefore i use 8C1
16 text colours there, therefore i use 16C1
result is 8C1 x 16C1 = 128
sad case this is not the answer, i refer to the answer, answer is 120
so any mistakes there?

Billy Bob
#4
Sep15-09, 02:39 PM
P: 393
Permutation and combination

Quote Quote by look416 View Post
As what you have said,
8 background colours there , therefore i use 8C1
16 text colours there, therefore i use 16C1
result is 8C1 x 16C1 = 128
sad case this is not the answer, i refer to the answer, answer is 120
so any mistakes there?
Step 1: Number of background colours is correct.
Step 2: Suppose you used Colour #1 in step 1. You can't use that same colour in step 2. Depending on the colour you chose in step 1, your step 2 choices of colour are different. However, the number of choices in step 2 is always the same.
look416
#5
Sep15-09, 06:26 PM
P: 87
yes mate i got it
which mean i just have to expel the colour that i have chosen for the background
therefore is 8C1 x 15C1 which is equivalent to 120.
yeah thanks mate, you saved my day


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