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Permutations and Combinations

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VydekiE:
Hi,
Can I please get help on this Combination question
1. Four balls are simultaneously drawn from a bag containing three green and six blue marbles. Find how many drawings are possible if:
a) the balls may be of any colour
b) there are exactly two green balls
c) there are at least two green balls
d) there are more blue balls than green balls

thank you!!

RuiAce:
Let G denote a green ball and B denote a blue ball. Note that the question is worded so that the green balls are indistinguishable among each other, as are the blues.

For a), if we consider any four balls drawn
_ _ _ _
They can either be blue or green. So 2 options for either of the 4 spots.
But G G G G is not possible, because we don't have 4 green marbles.
So the answer is 2^4 - 1 (15).

For b), because there are EXACTLY two green balls, we are essentially arranging:
B B G G
Which can be done in 4!/(2!2!) ways. Note that we're arranging because these 4 letter 'words' correspond to the possible arrangements of the orders they're drawn in.

For c), if there are at least two green, we must also consider when there are three green balls
B G G G
Which can be arranged in 4!/3! ways.
So the answer is 4!/(2!2!) + 4!/3! <--Edit, fixed a typo

d) note that this is ONLY possible if there are 3 or 4 blue bald
3 blue balls: 4!/3!
4 blue balls: only 1 possible outcome
So just add those together

VydekiE:
Thank you so much!!  :)

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