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wb47

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In a bag of marbles there are:
4R 3Y 2G
How many different arrangements of just 5 marbles are possible if the section must contain all 3 colours?

Is there an easier way than doing
RRRYG RRYYG RYYYG RYYGG RRGGY and adding 5!/3! + 5!/2!x2! etc etc?

Also I have no clue on how to work on this question:

A security password is made up by pressing 2 letters (out of ABCDEFGH) followed by 2 numbers (out of 1234)
i) how many diff. passwords can be arranged if letters and numbers CAN be repeated?
ii) Cannot be repeated.

Thanks!
 
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math man

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for the first one there really isnt anything better you can do..other methods would be to work out total number of combinations minus all the cases we dont want, but that is pretty long for this one too.

For the second question there are 8 letters and 4 numbers...we want to pick two letters followed by two numbers, this can be done as follows:

First letter can be picked in 8 ways
Second letter can be picked in 8 ways too as there is repetition
First number picked in 4 ways and second number picked in 4 ways too.

Therefore total passwords= 8x8x4x4

Without repetition both second picks will be one less as there will be one less option for the 2nd pick,

Hence (without replacement) no. pw's = 8x7x4x3
 
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TheCardician

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For the security password question:

i) You can have 2 letters out of 8 letters, and 2 numbers out of 4 numbers. Since the numbers and letters can be repeated: 8 x 8 x 4 x 4 = 1024 different passwords.
ii) If they cannot be repeated: 8 x (8-1) x 4 x (4-1) = 8 x 7 x 4 x 3 = 672 diff. passwords.

Hope this helps :) And for the marble question, I'm not so sure.

Edit: Damn Math man beat me to it xD
 

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