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26 Nov 2005, 13:27
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From 4 oranges, 3 bananas and 2 apples, how many selections of 5 pieces of fruit can be made, talking at least one of each kind.
31 Aug 2019, 02:13
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sandman
Please post a screenshot of the exact question. I am not sure whether the fruits are to be taken as identical or distinct e.g. whether the 4 oranges are identical or distinct. If nothing is given, I would be tempted to take them as identical.
In that case, I would split 5 into 3 groups with at least one each in the following ways:
{3, 1, 1} - 2 ways (3 oranges or 3 bananas)
{2, 2, 1} - 3 ways (1 orange or 1 banana or 1 apple)
So 5 ways.
But if the fruit pieces are distinct, then we have 9 distinct pieces of which we have to select 5 in 9C5 ways.
Out of these 9C5 ways, you are not allowed those in which one type of fruit is not taken.
No of ways in which oranges are not taken = 5C5
No of ways in which bananas are not taken = 6C5
No of ways in which apples are not taken = 7C5
So total acceptable ways = 9C5 - 5C5 - 6C5 - 7C5 = 126 - 1 - 6 - 21 = 98
General Discussion
26 Nov 2005, 23:03
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total combinations = (a + b + 3o), (a + 2b + 2o), (a + 3b + o), (2a + b + 2o), (2a + 2b + 1o)
possible numbers of combinations = (2c1 x 3c1 x 4c3) + (2c1 x 3c2 x 4c2) + (2c1 x 3c3 x 4c1) + (2c2 x 3c1 x 4c2) + (2c2 x 3c2 x 4c1)
= (2x3x4) + (2x3x6) + (2x1x4) + (1x3x6) + (1x3x4)
=24 + 36 + 8 + 18 + 12
= 98
possible numbers of combinations = (2c1 x 3c1 x 4c3) + (2c1 x 3c2 x 4c2) + (2c1 x 3c3 x 4c1) + (2c2 x 3c1 x 4c2) + (2c2 x 3c2 x 4c1)
= (2x3x4) + (2x3x6) + (2x1x4) + (1x3x6) + (1x3x4)
=24 + 36 + 8 + 18 + 12
= 98
