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26 Oct 2005, 20:06
Kudos
Bookmarks
I donno which thread should this go in.
There are 9 fruits. (could be of different types and/or sizes)
Question: In how many different ways can you form a fruit basket with
atleast 1 fruit in the basket?
Solution: Outcomes: Selected / not selected = 2
Repeated occurances = 9
Therefore possible outcomes after 9 repeated occurances = 2**9
which includes the one with no fruits seleted
Therefore no. of fruit combos = 2**9 -1
Is this correct ? or is 2**9 correct as mentioned in one of the earlier
threads?
Now consider that there are 3 apples, 4 plums and 2 grapefruits (each
could
be of different size/type)
Atleast one of each type should be in the basket
How many different combos are possible
Answer = 2**3 -1 X 2**4 -1 X 2**2 -1 (using the (No. of
oucomes)**repeated occurances formula and then subtracting 1 from each representing the outcome none is selected) [ No. of outcomes = 2
selected/not selected, repeated occurances = no. of fruits of each
type)
or
Answer = 2**(3-1) x 2**(4-1) x 2**(2-1) this assumes 1 of each is
already in the basket (implied here is the fact that no regard is being paid to the fact that the 3 apples may not be the same, machintosh, gala, red
delicious)
then this method assumes that out of the remaining two apples , 3 plums
and 1 grape fruit there is a difference in type/shape.
Which one is 'more' correct?
There are 9 fruits. (could be of different types and/or sizes)
Question: In how many different ways can you form a fruit basket with
atleast 1 fruit in the basket?
Solution: Outcomes: Selected / not selected = 2
Repeated occurances = 9
Therefore possible outcomes after 9 repeated occurances = 2**9
which includes the one with no fruits seleted
Therefore no. of fruit combos = 2**9 -1
Is this correct ? or is 2**9 correct as mentioned in one of the earlier
threads?
Now consider that there are 3 apples, 4 plums and 2 grapefruits (each
could
be of different size/type)
Atleast one of each type should be in the basket
How many different combos are possible
Answer = 2**3 -1 X 2**4 -1 X 2**2 -1 (using the (No. of
oucomes)**repeated occurances formula and then subtracting 1 from each representing the outcome none is selected) [ No. of outcomes = 2
selected/not selected, repeated occurances = no. of fruits of each
type)
or
Answer = 2**(3-1) x 2**(4-1) x 2**(2-1) this assumes 1 of each is
already in the basket (implied here is the fact that no regard is being paid to the fact that the 3 apples may not be the same, machintosh, gala, red
delicious)
then this method assumes that out of the remaining two apples , 3 plums
and 1 grape fruit there is a difference in type/shape.
Which one is 'more' correct?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
26 Oct 2005, 21:32
Kudos
Bookmarks
the answer would be
9C1 + 9C2 + 9C3 + 9C4 + 9C5 + 9C6 + 9C7 + 9C8 + 9C9
That is equal to (2^9) - 1 as you said
9C1 + 9C2 + 9C3 + 9C4 + 9C5 + 9C6 + 9C7 + 9C8 + 9C9
That is equal to (2^9) - 1 as you said
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
